为什么将字符串存储在字符中时只显示一个字符？

Question

[Im making a random word generator, so i use time.h library to use srand and generate a random number... If random number = 0, then char (principal)= 'hola'. But in CMD only shows the final character :( (Sorry for my bad english)

The code is:

#include <iostream>
#include <stdlib.h>
#include <time.h>

using namespace std;

int main()
{
    int principal;
    char principal1;

    srand(time(NULL));

    principal = rand() % 2;

    if (principal == 0)
    {
        principal1 = 'hola';
        cout << principal1 << endl;
    }
    system("pause");
    return 0;
}

Answer 1

In C and C++, a single quote (') means "character" (or one letter/number/symbol). Only one character should be between two single quotes. Example:

'c'
'9'
'?'
'\n'  // Newline character
'\0'  // Null character

A double quote (") means "string". A string literal (a string you typed into the code between quotes) will be treated as an array of characters with a null character added on the end. This type of character array is called a "c string" (Not to be confused with Microsoft's " CString ".

"Hi"  // Same as {'H', 'i', '\0'}
"Hello !\n"  // Same as {'H', 'e', 'l', 'l', 'o', ' ', '!', '\n', '\0'}

Trying to put a string inside single quotes is generally not good. It puts you into some weird territory of implemenation (ie compiler) specific behavior.

If you want to handle the string "hola", then you should use a std::string object to store it. (You could use a char* , but I recommend using std::string until you understand the differences better.)