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How do you read the line through the scanf?

T
terrea last edited by
I'm putting the line through the index.
```
int main(){
    char *s;
    scanf("%s", &s);
    printf("%s", s);
}
```
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1 Reply Last reply

magpie QA Engineer

We need to celebrate where to read. ♪ ♪

char * s = malloc(1024); // Или сколько вам надо
scanf("%s",s);      // Чтение ОДНОГО СЛОВА
scanf("%[^\n]",s);  // Чтение строки до конца (символа новой строки)

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How to create and fill a matrix according to the reading of each line of a file?
Software Programming • python functions numpy arc languagehive matriz • • yajahira

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C

It is unclear what to do with lines of more than 30 words. Since it is mentioned that if there are less than 30 words they must be filled with spaces, I assume that the M matrix must have mandatory 30 columns (and therefore if in one line there are more than 30 words you have to stay alone with the first 30 and ignore the rest).Your problem is that you are creating a new M in every loop iteration, and filling it with what you have read in the last line. When you leave the loop you will only have repeated copies of that last line.The matrix M it should be created before entering the loop that fills it. The problem is that at that point we still don't know how many lines the file has.Therefore I think the easiest thing would be to start reading the file to a series of lists (a list whose elements are in turn ready, each element of the external list represents a line and each element of the inner one the words of that line), then create the M matrix (now we will know how many lines it has, and the columns are 30) and finally in another loop go through words from the list intally read to the matrix.For example:import numpy as np def cargarArchivo(nombre): lista = [] Leer inicialmente el fichero with open(nombre,"r") as f: for linea in f: lista.append(linea.strip().split()) Crear la matriz filas = len(lista) M = np.empty((filas, 30), dtype="U20") Rellenar la matriz con lo leido del fichero for i in range(filas): for j in range(min(30, len(lista[i]))): M [i,j] = lista[i][j] return M By creating the matrix you already automatically have all your pre-initialized inputs with empty chains, so we don't need to add these if the lines are shorter. The trick is on for j in range(min(30, len(lista[i]))). Function min will give you the least of these two numbers: 30 or the number of words on the line. If the line has for example 10 words, the result will be 10, so I know j will vary only between 0 and 9 (leaving in the rest of the M the empty chains that already existed). If instead the line has for example 40 words, the result min() It will be 30, so j It will vary only between 0 and 29, thus avoiding the size of the matrix (ignoring the words that surpass in the line).
A

Delete line jumps within a list, read a file (name, phone number) and I must enter that information into a dictionary
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R

There are many things to tell about the code you have included in the question. Let's take steps:You have a function to read the file. As you have written what returns is a list, whose elements would be the linesYou have a function that takes a list of chains and concatens, to turn everything into a long unique chain.Calls on both functions text(read())Which doesn't make much sense. Why split the file into lines within read() to then rejoin all in one chain within text()? For that you better read the file read() instead of with readlines()so you get it in a single chain, and you save the call to text()When dividing the text in words you use .split(" "), passing a space as a parameter. This works only if each word is separated from the next by a space and only one. But it will give you problems if the separator are several spaces, or a tabulator, or an invisible character. It's better to call. .split() without parameters to work also in the previous cases. Besides, it turns out that .split() without parameters also consider \n as a separator, so it will work with words at the end of the line, and you will prevent those \n appear in the result.To not have to walk by opening and closing the "hand" file (i.e., using .close()) it is recommended to use a context (sentence) with). This way you don't need to close the file because python does it for you when the context ends.You don't need to pass the option "r" opening the file, since that is the default mode.Together with all the above, this would be my solution:with open("Names.txt") as f: contenido = f.read() data = contenido.split() print(data)
J

Why don't you work 7z.exe when you're hanging through spawnlp or execlp?
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I think we should call it.spawnlp(P_WAIT, "7z", "7z", "t", "C:\\Users\\Administrator\\root.war", NULL); The path to the programme and the zero argument of the command line are passed separately and have the right not to coincide. (Trust me that you don't need to indicate the full path.)
E

Read the file and record the data in a line
Software Programming • golang os • • elysha

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Example https://play.golang.org/p/jEITDk9fkt package main import ( "bufio" "bytes" "fmt" "log" ) func main() { lines := <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd"> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8"> <title>Title</title> </head> <body> Very fast! </body> </html> file := bytes.NewReader([]byte(lines)) // defer close real file scanner := bufio.NewScanner(file) c := 1 for scanner.Scan() { var s string if c == 8 { s = "Hello " + scanner.Text() } else { s = scanner.Text() } // записывать в новый файл построчно // меняйте os.Stdout на файл, буфер и тд и тп fmt.Fprintln(os.Stdout, s) c++ } if err := scanner.Err(); err != nil { log.Println(err) } }
B

Why don't you want to split my program?
Software Programming • c++ c language • • baileigh

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A

If you do this PROMEDIO = CAL1 + CAL2 + CAL3 + CAL4 + CAL5 / 5; is equivalent to this: PROMEDIO = (CAL1 + CAL2 + CAL3 + CAL4) + (CAL5 / 5); So to solve your problem what you have to do is put parenthesis to make you more accurate and put it like this:PROMEDIO = (CAL1 + CAL2 + CAL3 + CAL4 + CAL5) / 5;
T

I can't read a line file by line
Software Programming • java • • terrea

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J

To iterar about the array uses its length you get through the property length ... String array[] = bfRead.split(" "); for (int i = 0; i < array.length; i++) { ... ... ... } To read the lines using BufferedReader must be done in this way:String line = br.readLine(); while (line != null) { System.out.println(line); line = br.readLine(); } Implementing the above to your method would be: public static String LeerTxt(String direccion){ String texto = ""; try{ BufferedReader bf = new BufferedReader(new FileReader(direccion)); String bfRead = bf.readLine(); while(bfRead != null){ String array [] = bfRead.split(" "); for (int i = 0; i < array.length; i++) { if(array[i].isEmpty()){ //No tiene elementos }else { texto += array[i].trim()+";"; } } bfRead = bf.readLine(); } }catch (IOException e) { System.err.println("IOException " + e.getMessage()); }catch (Exception e) { System.err.println("Exception " + e.getMessage()); } return texto; }
K

How can I read or print a block of lines or data from a file, between two lines or two keywords within the file on python?
Software Programming • python python 2.7 • • Kadyn

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Let's assume that the start and end chains have them in the variables inicio and finalrespectively. It could be done this way:inicio = "Datos del archivo" final = "Final de los datos" resultado = [] with open("Archivo.txt") as fichero: # descartar las primeras líneas fpos = (linea for linea in fichero if not line.startswith(inicio)) # recoger las líneas hasta la siguiente marca for linea in fpos: if linea.startswith(final): break resultado += linea In a somewhat more elegant and functional way:from itertools import takewhile inicio = "Datos del archivo" final = "Final de los datos" with open("Archivo.txt") as fichero: # descartar las primeras líneas fpos = (linea for linea in fichero if not line.startswith(inicio)) # recoger las líneas hasta la siguiente marca resultado = list(takewhile(lambda s: not s.startswith(final), fpos))
N

when tightening enter, the cmd opens lock even by completing all commands
Software Programming • c language • • Nykeriab

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D

You are forgetting to make use of '&' in the scanf......, &valor1) ... ..., &valor2) ... ps: also has a wrong formatting on your second printf(), just replace the char $ per %.
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How do you read a few numbers in one line when the number of numbers in the line increases by 1 per Python 3.7 each time?
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blist = [] for i in range(val): x = input().split() blist.append(list(map(int, x))) by split We share the gap line. map translated int each part of the list
R

How is using acces() opening a security hole?
Security, Penetration Testing • c language linux • • rielyn

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R

It is a race condition. You do the access(), then you do the open(). In the small time between the two calls, the file may have changed. Typically, the file is, say, /tmp/foo. Initially, the file is owned by some user (who is the bad guy of the story), and the target is some root-powered application. The application does the access(), sees that the file really belongs to the user, and thus thinks: "that's fine, it's his file, I can process it on his behalf". Then the bad guy quickly replaces the file with a symbolic link to /etc/shadow. The application has already taken the decision to open /tmp/foo, but when it does, it really opens and processes /etc/shadow.
Y

Create new lines from the line
Software Programming • javascript • • yajahira

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var string = "Регулярные выражения - это шаблоны используемые для сопоставления последовательностей символов в строках."; console.log(string.split(' ').join('\n')); // или раз уж зашла речь о регулярках: console.log(string.replace(/\s+/g, '\n'));
J

How to read only specific lines of .txt files in C
Software Programming • c language file • • juvenalb

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M

You should use files in binary mode, they have similar operation, to create/read/write in binary files you should change the opening parameters of the fopen(), e.g. fopen("arquivo.txt", "rb");. Using files in binary mode just use the function fseek().
F

How do we reduce the search time in the rig (more)?
Software Programming • c language linux algorithm • • fbio

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If time is critical, when the data are downloaded, sort a bunch of tags in any slow way (short of Shell, for example), then search the row in the area with a binary search.For example:long int searchBinary(dio_table *t, const requestData *rdata) //t - структура с массивом указателей на строки и их количеством, rdata - данные для поиска { lint bottom = 0; // Индекс начала поиска lint mid = 0; lint top = t->head.count - 1; //Индекс конца поиска while(bottom <= top){ mid = (bottom + top)/2; int cmpr = strcmp (t->lines[mid].data, rdata->col); //t->lines[mid].data получаем строку по середине, сравниваем с rdata->col if (cmpr == 0){ // Если нашли искомую строку, то ее и возвращаем return mid; } else if (cmpr > 0){ // Если сравнение вернуло положительное число, значит искомая строка где-то выше по списку top = mid - 1; } else if (cmpr < 0){ // Или ниже bottom = mid + 1; } } return -1; // Если ничего не нашли } Like that. But again,Такой поиск работает ТОЛЬКО в отсортированном массиве! UPD: the volume is to be sorted at an increase
How do you compare the three values in si?
Software Programming • c language algorithm comparison • • jeanid

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C++#include <iostream> using namespace std; int main(int argc, char **argv) { int a = 5, b = 2, c = 8; cout << min(min(a, b), c) << endl; return 0; } C#include <stdio.h> int min(int x, int y) { return x < y ? x : y; } int main(int argc, char **argv) { int a = 5, b = 2, c = 8; printf("%d\n", min(a, min(b, c))); return 0; }
K

How to let the string size be set by scanf()?
Software Programming • c language string memory management sc languageanf • • kalena

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Actually the question part of the wrong premise. The code that creates string is limiting in 5 characters, has nothing unlimited there. This code will book 6 bytes in the code to allocate the 5 characters plus the terminator, and at the time the function is called will be allocated a pointer to that region. You can't change that size.What can be done is create another string completely different, allocated in the static part of the code like that of the code, allocated in the heap dynamically or automatically stack, and the address where this new string can be placed in the variable teste. https://pt.stackoverflow.com/q/106719/101 .O https://pt.stackoverflow.com/q/168882/101 , that's even one of the criticisms you make to it. It just doesn't mean that memory won't be corrupted.What you can do is read character by character and go relocating memory. I made a https://pt.stackoverflow.com/q/177409/101 . An improvement there would be going folding of size each relocation and already starting with a reasonable size, this would make the logarithmic complexity. That is, write your own scanf().Read https://pt.stackoverflow.com/q/3797/101 .

How do you read the line through the scanf?

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