Why just, if it's hard to... ♪vector<int> v; int i; . . . while(file >> i) { v.push_back(i); } If the file is open out for recording:bool wasNeg = false; int i; while(file >> i) { if (!wasNeg && i < 0) { wasNeg = true; continue; } out << i << " "; } And... Don't do that.while (!file.eof()) This test will only work after failed Reading. Better like this:while(file.getline(...))

So look at the prototype:int atoi(const char *nptr); The function is taken by the index on the line (char *and you put a symbol on her.char) That's why they don't.And to be one symbol (subject to which it is a decimal figure) very simple:int two = str[1] - '0';

Under ubuntu, the table of default symbols utf8. Its symbols above 127 should be at least two-bite.

Problem solved strangely. During the debaga, why did I see a mass in the console that I removed a long time ago (I don't know what he was doing there, because he wasn't on the code). After that, because I went through the phone, I pulled the app, started over, and it all worked.P.S. He was in Android Studio.

I'll write a synonymous version, since it's probably a seal in the field, and it's impossible to do it with one length.public static int pos(String search, String where) { //искомая строка должна быть не больше той, в которой ищем if (search.length() > where.length()) { return -1; } //ищем только до определенной позиции, т.к. дальше искомая строка уже не "влезет" for (int i = 0; i < where.length() - search.length() + 1; i++) { // если совпадает первый символ ... if (where.charAt(i) == search.charAt(0)) { boolean found = true; // ... то сверим все остальные for (int j = i + 1; j < i + search.length(); j++) { // если хоть один не совпадает ... if (where.charAt(j) != search.charAt(j - i)) { // ... значит не нашли found = false; break; } } // если все символы совпали if (found) { //значит нашли и возвращаем позицию return i; } } } //ничего не нашли return -1; }

You don't understand that you want to answer you seriously.Here you go:unsigned long s = 'test'; printf("%lX\n",s); Conclusion 74657374 And there's an idea. test 16.Just don't show me... :

One. The type of bike in C is called uint8_t♪ Lives stdint.hTwo. Use it. fwrite(str,sizeof(str),1,stdout);

https://stackoverflow.com/questions/7019454/why-does-stdcout-output-disappear-completely-after-null-is-sent-to-it ANNOUNCEMENTConclusion char* = nullptr assistance cout Call. Undetermined behaviour♪ It's a violation of the language standard, and you can't do it.In case libstdc, it's "Indistinct behavior" that becomes quite certain, things turn around like this:iostreams Trying to interpret char* Like a line. And when you make him zero, he's messing up, he's making mistakes and he's not working.A little experiment on http://ideone.com/9r7y9m :#include <iostream> #include <cstdio> int main() { char* pC = nullptr; //использование nullptr //вместо NULL - использование //современного стандарта языка std::cout << pC; // выводим статус потока, будет "0" printf(" goodbit state is %d\n",\ std::cout.rdstate() == std::ios_base::goodbit); std::cout &lt;&lt; "Any output"; // не сработает - cout сломан std::cout.clear(); //сбрасываем ошибку std::cout &lt;&lt; "Any output"; //теперь вывод работает return 0; } But it's not worth it. It's the right thing not to try to make. nullptr♪If you have to put the sign on the line, bring his type to something else. std::cout << static_cast<void*>(pC) << std::endl;

First option:сhar s[] = { 0x05, 0x06, 0xD1, 0xE0, 0xEC, 0xE0, 0xF0, 0xE0 };` keeps code sequences. The total is 8 bytes.Second option:std::string s = "0x05, 0x06, 0xD1, 0xE0, 0xEC, 0xE0, 0xF0, 0xE0"; keeps the line in a clear-cut form. The total size is 47 Byte (if my calculator is not wrong). The second option would include:Dividing underground (coloured);Transfer from line to number, e.g. through http://www.cplusplus.com/reference/string/stoi/ ;Fill a new set of numbers.In general, the answer to the question is:How am I supposed to fill Char s[s] from the variable?Depends on how this "replaceable" is stored.If you just have to put the line in code, then write the codes, not the lines:std::string s = "\x05\x06\xD1\xE0\xEC\xE0\xF0\xE0";

use http://www.cplusplus.com/reference/cstdlib/mbstowcs/ In the code, it'll be about that.#include <cstdlib> #include <cwchar> #include <memory> #include <string> std::wstring to_wstring(const char* str) { std::unique_ptr<wchar_t[]> tmp = nullptr; size_t sz, len; len = mbstowcs(nullptr, str, 0); // получить размер sz = len + 1; tmp.reset(new wchar_t[sz]); // выделить память mbstowcs(tmp.get(), str, sz); // перекодировать return std::wstring(tmp.get()); } std::wstring to_wstring(const std::string& str) { return to_wstring(str.c_str()); } bool _IsProcessRun( const char * processName ) { HANDLE hSnapshot = CreateToolhelp32Snapshot( TH32CS_SNAPPROCESS, 0 ); PROCESSENTRY32 pe; pe.dwSize = sizeof( PROCESSENTRY32 ); Process32First( hSnapshot, &pe ); std::wstring name = to_wstring(processName); while( Process32Next( hSnapshot, &pe ) ) { if( name == pe.szExeFile ) return true; } return false; }

You can still be like this through. union:typedef union { char v; struct { unsigned char c: 4; unsigned char i: 4; }; } value; int main(int argc, const char * argv[]) { value v; v.v = 0x35; cout << char('A' + v.c) << endl; cout << int(v.i) << endl; }

Where strlist It is guaranteed that it ceases to be used, but as soon as possible (to the point where memory has been provided).

Indices in a mass of three elements may be between 0 and 2.You write beyond the limits of the distributed memory:char* tempm = new char[3]; ... tempm[3] = '\0'; et al.I mean, that's it. temph[3] = '\0'; Probably washing the place where it starts. tempm♪

The comparison of the two lines in C is carried out as follows:#include <stdio.h> #include <string.h> int main() { char a[100], b[100]; printf("Enter the first string\n"); fgets(a, sizeof(a), stdin); printf("Enter the second string\n"); fgets(b, sizeof(b), stdin); if (strcmp(a,b) == 0) printf("Entered strings are equal.\n"); else printf("Entered strings are not equal.\n"); }

Use it. https://docs.oracle.com/javase/8/docs/api/java/util/Iterator.html ♪ Implementation outline:class ReaderIterator implements Iterator<Character> { private final Reader reader; private int next = -1; public ReaderIterator(Reader reader) { this.reader = reader; } @Override public boolean hasNext() { if (next != -1) { return true; } else { try { next = reader.read(); return (next != -1); } catch (IOException e) { throw new ReadException(e); } } } @Override public Character next() { if (!hasNext()) { throw new NoSuchElementException(); } Character result = Character.valueOf((char)next); next = -1; return result; } } Then send the terator to the steamer:fileRead = new BufferedReader(new FileReader(pathFile)); ReaderIterator iterator = new ReaderIterator(fileRead); parser.Parse(iterator); // Parse(Iterator<Character> iterator) There are symbols in the password:while (iterator.hasNext()) { char c = iterator.next().charValue(); // ... } Insufficient ReadException must be unchecked.

It's all about syntaxis."+" - it's a line.Stringwith one symbol +'+' - it's a symbol.char) +Since str_char[i] Type char♪ switch You should use the second option.

To this end, the standard needs to add the withdrawal flows for these types. At this point, there is no such opportunity in C+++. At present, if the console supports utf-8, the reference symbols in utf-8 can be converted and removed.

It's just that the author of the algorithm didn't want to encrypt and decode the algorithm of the fashion degree (this algorithm used in the encryption and decoding of the algorithm description) and went through that mode.long long modular_pow(long long base, unsigned long long exp, unsigned long long modul) { long long d = 1; unsigned long long mask = 0x8000000000000000; long long modulus = modul; unsigned int shift = 63; for (unsigned int i = 0; i < 64; ++i) { d = (d*d) % modulus; if ((exp&mask) >> shift) d = (d*base) % modulus; mask >>= 1; --shift; } return d; }

If length mass in line form,#include <string.h> char charArr[] = "string"; strlen(charArr); // 6, размер самой строки strlen(charArr) + 1; // 7, размер массива с символом окончания строки UPD:If length mass int array[]:return sizeof(array)/sizeof(array); If it is. Dynamic massand There is no end-of-life criterionThere's no answer.

http://dev.mysql.com/doc/refman/5.7/en/storage-requirements.html ♪ Char, varchar, enum, text and others are limited in length in symbols. But the general limits (e.g. length of the line in the table, or size of the index) may be further limited in the whites.The direct text is written in http://dev.mysql.com/doc/refman/5.7/en/charset-unicode-utf8.html :For example, MySQL must reserve 30 bytes for a CHAR(10) CHARACTER SET utf8 column.char(10) in utf8_general_ci will borrow 30 bytes, regardless of the data recorded.