c++ - srand() 的问题，C++

Question

我正在尝试编写一个使用种子生成伪随机数的程序。但是，我遇到了问题。

我收到这个错误

39 C:\Dev-Cpp\srand_prg.cpp void value not ignored as it ought to be 

使用这段代码

#include <iostream>
#include <iomanip>
#include <sstream> 
#include <limits>
#include <stdio.h>

using namespace std ;

int main(){
    int rand_int;
    string close ;

    close == "y" ;

    cout << endl << endl ;
    cout << "\t ___________________________________" << endl ;
    cout << "\t|                                   |" << endl ;
    cout << "\t|   Pseudorandom Number Game!       |" << endl ;
    cout << "\t|___________________________________|" << endl ;
    cout << endl << endl ;

    while ( close != "y" ){

        rand_int = srand(9);
        cout << rand_int << endl ;

        cout << "  Do you wish to exit the program? [y/n]     " ;
        cin >> close ; }

}

Answer 1

srand 不返回随机数，它只是重新播种随机数生成器。之后调用 rand 实际得到一个数字:

srand(9);
rand_int = rand();