It's not just struct. O typedef is used to define a type of data, as the name already indicates.In addition to the types already existing in the language you can create yours and like any other.A defined type can be something many simple, like just a different name (an abstraction) for an existing type, as may be the composition of some values in a given format that create a different form of value than it exists.You can do this with the struct, but if you only do with it does not create a new type for language to use with an available name whenever you upload it. It will still create a new type, but it will be considered anonymous for the language, it will have a name only for the command struct, which is boring to use and can cause some misconceptions. With typedef everything gets easier because it leaves simple, clean and correct. Basically he's an abstraction.It is usually recommended to use the capital at the beginning of the name to avoid conflicting with other symbols, or to use a suffix _t. Something like this:typedef struct { int x; int y; } Ponto; Here you use in the statement of a variable:Ponto ponto; Or cast:(Ponto)ponto Without it:struct ponto ponto; Or:(struct ponto)ponto https://github.com/maniero/SOpt/blob/master/C/Typing/Typedef.c .When you use a name struct the name pattern is less important because it is certainly a type, it cannot be used, and therefore confused, with other code identifiers. When you use one typedef a struct You don't have to have a name. You may have one struct anonymous without typedef, but there it cannot be used anywhere else in the application.

The problem is that you are using the variable before bootingint opcao,codigo,quantVend,y; //Cria a variável y produto roupas[y]; //Usa a variável y y = 3; //Atribui um valor a variável y The order must be thatint opcao,codigo,quantVend,y; //Cria a variável y y = 3; //Atribui um valor a variável y produto roupas[y]; //Usa a variável y

I decided to help because it has an extraordinary amount of errors in the code. The ideal is to learn one thing at a time, so make only one mistake, learn from it, move on to the next and there will evolve knowledge. When you try to do what is not yet prepared you make many mistakes and do not learn from them. One of the mistakes and not put the code more organized to help better. Programming is a lot of organization, who cannot do this will have difficulties in the profession.#include <stdio.h> typedef struct { int id; char nome[201]; char endereco[101]; char cidade [31]; } Pessoa; void preencherPessoas(Pessoa pessoas[], int qtde) { for (int i = 0; i < qtde; i++) { printf(" Digite o ID do funcionário: "); scanf("%d", &pessoas[i].id); printf(" Digite a nome do funcionário: "); scanf("%s", pessoas[i].nome); printf(" Digite o endereco do funcionário: "); scanf("%s", pessoas[i].endereco); printf(" Digite o cidade do funcionário: "); scanf("%s", pessoas[i].cidade); printf(" DADOS DO FUNCIONARIO DE ID %d CADASTRADOS! \n", pessoas[i].id); } } int main() { printf("Insira a quantidade de funcionários que serão cadastrados: "); int qtde; scanf("%d", &qtde); printf("\nQTD: %d", qtde); Pessoa pessoas[qtde]; preencherPessoas(pessoas, qtde); } See https://ideone.com/bRJIc6 . And https://repl.it/join/sgsffaia-maniero . Also https://github.com/maniero/SOpt/blob/master/C/Function/ArrayPass.c .I gave a name that means something in the kind of structure created. Code is expression, it is important to give good names to things to better understand what you are doing. Every experienced programmer does that, only novice gives no value to this.I put another 1 byte in strings to fit https://pt.stackoverflow.com/q/177619/101 . The sizes are absurdly large and waste a lot of space and any amount of larger data will burst the stack. I'm not going to go into detail, but all that idea of organizing the struct It's wrong and real codes aren't like that. It serves only to understand other things in this little code, keep it in mind.Do not declare global variables unless you have a very large domain of what you are doing, just declare local variables.Give a better name for the function. What would happen if I had to fill something else?If you created a guy then use it not using the struct to declare variables and parameters.No need to pass array by reference because it is already a reference. I would only need to pass this way if I were to change the whole object, which is not the case, even you can't even do it in the planned form.Do not need to declare the variable before its use, in fact this makes the code less readableNames using ALL_CAPS notation should only be used for macros, and still should be avoided. And you don't need to abbreviate so much, use a more common abbreviation.A function that must fill people's data should not ask how many people will be in total. This is the responsibility of another function.If the data is a string then he's already a array, therefore it is already a reference and scanf() asks for a reference, so you do not need to create another reference when you pass this data, it is already naturally what you expect.I created a main() that makes sense.If you didn't understand anything, it's because you jumped too far. Here on the site has almost everything you need to know before continuing.

If you just want to save in a variable, you can use sprintf in place of printf:#include <Windows.h> #include <stdio.h> #include <time.h> void main () { char minhaString[30]; time_t t = time(NULL); struct tm tm = *localtime(&t); sprintf( minhaString, "Agora: %d/%d/%d", tm.tm_mday, tm.tm_mon+1, tm.tm_year+1900 ); }

The int i starts at 0 in a 1 stool, you're not using the first vector position.This is in the fread manpage: On success, fread() and fwrite() return the number of items read or written. This number equals the number of bytes transferred only when size is 1. Maybe yours while condition is wrong, try to put the > 1It might work out.I hope he helped you.

They can't, return and goto. Only sizeof there is an operator, and yet it has a special way, because it always runs in compilation time. It's not the only one who can link as an operator. Then the question part of the wrong premise.These are commands statements of language, are reserved words, such as sizeof.union and struct are statements also, but they have a more declarative function, they are not exactly imperative as the previous ones. They determine that there will be a declaration of a data structure of union or structure composed of data. They declare new types for their code. So they're not operators.Operators have the function to manipulate (access) data and result in something, are expressions.More: https://pt.stackoverflow.com/q/173379/101 https://pt.stackoverflow.com/q/168177/101

You have a lot of problems in that code and even the question, but I'll focus on what you asked.This free() doesn't do what you're wondering. You can only give free() in something you gave malloc(), exactly the same thing, cannot invent something else to give free(). You could just not have put on the question malloc() in the name, but I doubt it, and if you did it, it's wrong.If nome is a given inline in the structure does not have what to allocate outside of it, so do not have why to give a malloc() for him, what I don't think you did. The data is already inside struct, If you allocate space for it, the whole space is already allocated, and if you need to release the space, it has to be done for all struct.If you want to have the name with variable size and outside struct then put a pointer there just and then allocate the name on demand, but there is already another problem, and we do not have enough information to talk about it, actually this is even speculation.

Lista is a type to organize the list, it is not the next structure, it is the first, it is where the list begins. If all knot is accessed by means of a pointer, how is the first knot accessed? For an external pointer. That's all. There is not well a pointer for pointer, there is only one pointer for a structure, which inside has another pointer.

Because it is accessed through a pointer, then you need to disrefer the pointer before accessing the member.jose is a pointer for an object and not the object itself, it is a number with the address. To pick up the object itself has to say that you want it, using the operator * and then with the object you can take his member. Then you need to do (*jose).idade. Since this is quite common they have created a simpler syntax jose->idadeThat's the same thing.

Generally not, the correct thing is to pass everything as parameter, all that is global ends up getting lost in the organization. Of course, for something simple will not give a problem and can give the false sense of security, but when you start making real codes will get complicated, then learn to do the right way from the beginning. This does not depend on your problem. It is rare to need something global and need to have good programming domain to use properly. In doubt use parameters.

In its function List *CreateList () you allocate only one structure List that contains one int and a pointer for a Node, which is yours headAnd that's all. And after that you make your pointer point to NULL, therefore, in there has no structure representing a Node Indeed.So when you create the list, you have nothing but a structure with a int and an empty hand for a Node.Here's the thing:List *CreateList () { List *list = (List*) malloc(sizeof(List)); list-&gt;size = 0; //Aqui você está fazendo a variável head da estrutura recém alocada list //Aponta para NULL, já que você não alocou nenhuma estrutura para ela //apontar. Então, dentro do head não tem nenhuma estrutura Node de fato list-&gt;head = NULL; return list; } In function void Push the new Node alocado ta being positioned at the beginning of the list:void Push (List *list, DataNode data01) { Node *node = (Node*) malloc (sizeof(Node)); node-&gt;data = data01; //Aqui você está fazendo com que o ponteiro que está definido dentro da estrutura //Node, aponte para um endereço de memória, que é o endereço apontado //pela variável head da estrutura lista node-&gt;Next = list-&gt;head; //Aqui você está fazendo o ponteiro head da estrutura List apontar para o novo //Node alocado, que vai ser o novo head list-&gt;head = node; list-&gt;size++; } So when your list is empty and the function Push() be called for the first time node->Next will be equal to NULL, since the pointer list->head point NULL.So, to get easier to view, you are pushing all the elements of the list back and placing a new element at the beginning.Remembering that a pointer is a variable that you declare in some part of your code that will point to a memory address.

It is not actually possible to "inflate" a struct since this term defines a data structure or type of data. What you'll want to do is fill a container with struct type objects. Another thing... it'll be tricky to do what you want with this json file. The file defines an array of Items of three types, Product1, 2 and 3, I think it would be simpler:{ "Itens": [ { "id": 1, "tipo": 1, "valor": 20, }, { "id": 2, "tipo": 3, "valor": 33, }, { "id": 3, "tipo": 6, "valor": 60, } ] } If it can be this json then it becomes easy popular a vector-type container with objects of its struct type:#include <iostream> #include <fstream> #include <jsoncpp/json/json.h> using namespace std; struct Item{ int id, tipo, valor; }; int main() { vector<Item> containeritens; ifstream ifs("Arquivo.json"); Json::Reader reader; Json::Value obj; reader.parse(ifs, obj); const Json::Value& itens = obj["Itens"]; for (int i = 0; i < itens.size(); i++).{ Item item; item.id = itens[i]["id"].asUInt(); item.tipo = itens[i]["tipo"].asUInt(); item.valor = itens[i]["valor"].asUInt(); contairnerItens.pushback(item); } return 0; }

Rust allows you to boot an object with a literal shape. This form is composed by the type name, of keys that will indicate the beginning and end of this literal, as if it were the quotes of a string, and then you will place the members who want to boot the data of the object.Rust has no constructor like other languages, it can have simple functions that serve as builders, but the most widely used form is pure and simple startup in this literal form.So it looks like a function call with named arguments, only what you're doing is using the names of the members of the struct and their values. So what is before : is the name of the member and what is after the : is the value that will be assigned to that member.In this example it is a type Complex, re is the real part of the number and im is the imaginary part. Nothing has to do with what you speculated on the question, it's much simpler than this. This information has to do with the type of data there, Complex and not with rules of language. The language only determines the syntax of this object initializer.Another hypothetical example:let joao = Funcionario { nome : "João", salario : 1000 }; https://github.com/maniero/SOpt/blob/master/Rust/Field.rs .Do you understand? Just https://docs.rs/num-complex/0.1.42/num_complex/struct.Complex.html , and tell of passing poorly chosen names, in fact it looks like C and not Rust that usually abbreviate a lot.

The question could be considered very broad because it has several problems. Stack Overflow works best with well-defined issues in well-focused questions.I will list some issues of your code that you need to be corrected. From there you will be able to improve some things and be able to ask more focused questions when you have a more specific problem.The guy char stores only one character. To store multiple characters you need to declare as array, something like char nome[40] where you will book 40 characters inside your own struct which is well inefficient to can be used for didactic purposes. Or you can declare how char * nome and can have the indeterminate size, occupying only the space for a pointer (there will have to make the allocation out).You do not initialize the vector. So it is very easy to pick up a lot of garbage. The fact that you reserve memory for the variable in the statement does not mean that memory is clean.You are using an integer to save phone. Although a phone number seems a numerical value it is not. I would use one string for this too. Remembering to insist on using a int, you cannot use numbers more than 9 digits or start with zero (which technically does not exist).That's it. if ( escolha==3) inside default It's really weird. If you want to do something when the choice is 3, mount one case for that. And put everything there, you don't have to have the other if just below. It's weirder. while perform until escolha 5. And another detail, you just need one system ("cls"); in all your code.This way of checking if there is space in the vector to place new data is extremely flawed. You would need to boot the data with NULL and check this out.This variable espaco is completely unnecessary. If you find vacant space in the vector, you process the inclusion there in that if, do not need to create a confusing algorithm.If you are using a vector, you need to use an index in the variable. No scan probably you wanted to do &vetor[i]->nome. And the other probably &vetor[i].telefone (if you are using one int and not a pointer to int, would have to use the operator . and not the operator ->.Then the printf() is using the wrong operator again. You're printing the pointer when you want to print the content pointed out by it, so: &vetor[x]->nome.No printf() would have to use the specifiers %s and %d. The same mistake was made in scanf().You are using the variable x No loop but it does not vary, who is varying is the i.You can only say that the vector is empty if it passes through all for and no items are printed, you cannot say this while inside for.It has a series of small other details that will not make so much difference, as unused variables, or use MAX and now use a literal 3 that does not even match the current value of MAX, inconsistency between use of pointer and array, comments that say do something that is not real, etc. that make your code get weird or even wrong.I didn't do a very deep analysis. You may have more mistakes or I may have understood something wrong. You need to improve by parts.I think I've helped a lot, more than this just if you do for yourself and I'm sure you want to do it on your own.

The writing and reading in file you have is correct, the problem is some mistakes and misconceptions of yours in the code:As I said in comment printf is wrong because it prints a house that does not exist:printf("%d %c %c\n", read_data.partida, read_data.jogvelha[3][3], read_data.resultado); // ^--^ If you have a board of 3 by 3, then you have houses 0, 1 and 2. House three is out. It is important to remember that the first house of an array is always 0.Defined resultado as char in the structure:typedef struct { ... char resultado; } velha; But then it saves the number 1 value in this field:int ganhador=1; FUNCAO_QUE_GRAVA_BIN(local,partidas,ganhador); // ^-- aqui passa um int com 1 em vez de um char This will make you see such a strange caratere of the ASCII table, more in concrete the ASCII caratere 1.Booting jogvelha is also wrong:velha partida = {partida1, {"a","b"},resultado1} // ^^^^^^^^ If you have an array of chars 3 by 3 then you're missing characters, just a missing string. Another detail that you forgot is that if you save the values as strings then the compiler includes the terminator, which makes it less space, and needed to have a board 4 by 4. It is better to start as char even, writing the characters you want manually:velha partida = {partida1, {{'X','O','X'}, {'X','O','X'}, {'X','O','X'}}, resultado1};

Supposing you declared a t_agenda as the written below, to access an element of the vector contatos, you should do as follows://declaração da variavel t_agenda agenda; //acessa o dia do t_data, onde `indice` deve ser substituido pelo valor do indice agenda.contatos[indice].dataNascimento.dia;

First, you don't seem to need one union. https://pt.stackoverflow.com/q/47871/101 , which already kills the union need.union with two fields of the same kind doesn't make much sense. Just to use a different name? I don't see an advantage in C. In C++ it has better mechanisms. It's not what you need anyway.One union without Tag, one of the criticisms made to the C mechanism does not serve much, so C++ has better structures than union in the library, if necessary. But of course you can always use one Tag own extra in own struct to know what to catch in each case, only has the problem that this is an abstraction leak.union was created to solve lower-level issues, I think it is mixing different levels of abstraction.Therefore I prefer neither to try to solve your specific problem and solve the real problem that is wider. Solving the question problem is a palliative that will still remain wrong in the general concept.If you do it right you will only have the space for string of the CNPJ which is bigger, and of course, one Tag extra that seems fundamental to me in any way in this register. This is a case that puts string inline struct Looks appropriate.Question strongly if you should use a struct for one Cliente. But ok, it's just a start, and technically it will change little to use for a class, it will only give a more appropriate semantic, and indicate more that this should probably be a type per reference. But I'm not going to insist on it because right now it's too simple code.This would be a better solution:struct Cliente { string nome, email, tipo, documento; }; https://github.com/maniero/SOpt/blob/master/C/Struct/SimpleRecord.c .It has other issues that probably indicate that the code is mixing C with C++ and is the way of the last century to do in C++.

It's a lot of code issues.In function enfileirar() the parameter fim must be a pointer for pointer:void enfileirar(struct no **inicio, struct no **fim, int valor) {And that forces you to put the pointer fim in parentheses at the assignment:(*fim)->prox = novoNo;In function mostrarFila() no need to pass pointer to pointer since there will be no change in the listvoid mostrarFila(struct no *inicio) {And assignment aux is simple:aux = inicio;The test was wrong:while(aux != NULL) {And you want to print the data and not its address:printf("%d\n", aux->dado);Same thing in function main()printf("%d\n", inicio->dado);printf("%d\n", fim->dado);And call to mostrarFila() changes to:mostrarFila(inicio);Several of these problems were detected as warnings in the compilation.

The notation of used hands is not correct in some places, according to the declared type. When it does:*(*p).vet[1]="Ana"; You already have several problems, because vet is an array of strings, so vet[1] is a string, so you have one * the most at first. However assignment of strings in C is done with http://www.cplusplus.com/reference/cstring/strcpy/ and (*p). can be simplified to p->. Using all that I mentioned this instruction would be:strcpy(p->vet[1], "Ana"); At the pointer assignment q also has the same problem, and the statement of strings vet[100][4] certainly stayed the opposite because it ended up declaring 100 size strings 4.It ended up complicated enough in the solution even with the restriction of using pointers. If the goal is to reverse an array of 3 strings using pointers, then only this is enough:#include <stdio.h> #include <string.h> //função de inversão utilizando ponteiros void inverter(char *str){ int tam = strlen(str), i; for (i = 0; i < tam / 2; ++i){ char temp = *(str + i); *(str + i) = *(str + tam - i - 1); *(str + tam - i - 1) = temp; } } int main(){ char strings[4][100]; strcpy(strings[0], "Ana"); strcpy(strings[1], "Bia"); strcpy(strings[2], "Lia"); int i; for (i = 0; i &lt; 3; ++i){ inverter(strings[i]); printf("%s\n", strings[i]); } return 0; } https://ideone.com/eoxGC9

If you know how to control readings and how they work, you don't need to __fpurge or similar.In case it happens that reading with %[^\n] does not work if the first thing in the entrance is already one \n left by a forecast %d or %f for example. It has numerous ways to circumvent this problem but the simplest is to place a space before %[^\n] to consume the previous break.It is also wrong to use the char*, this:void Insere_elemento(PESSOA*dados) { char aux_nome[20]; scanf("%[^\n]", aux_nome); dados->nome=aux_nome; ... It is wrong because the string for the name was allocated in the function in stack and when this ends the space will be marked as free and so any access you make then represents indefinite behavior. The solution is to allocate the straightforward characters with http://www.cplusplus.com/reference/cstdlib/malloc/ and then do the reading for the allocated space.Example:void Insere_elemento(PESSOA*dados) { dados->nome = malloc(20 * sizeof(char)); //alocação com malloc dados->cargo = malloc(20 * sizeof(char)); //alocação com malloc scanf("%[^\n]19s", dados->nome); scanf("%d", &dados->idade); scanf("%f", &dados->salario); scanf(" %[^\n]19s", dados->cargo); // ^-- espaço consome a quebra anterior scanf("%f", &dados->tempo); } To note that the writing you are doing in Imprime_dados has the formattor %2.f when you probably wanted %.2f https://ideone.com/NqlOsf Also note that I forced the maximum reading size with scanf to 19 in order to avoid possible problems with buffer overflow.