在C中使用rand（）时为c4014

Question

I'm trying to make my program return a random number [-5,5] using rand() function in C. The program is working but I dont like the fact its giving 2 warning everytime i compile the code. Here is the code:

#include <time.h>
#include<stdio.h>
int main(int argc, char *argv[]){

    int randomNumber = 0;
    for (int index = 1; index < argc; index++) {
        printf("The %d is %s\n", index, argv[index]);
    }
    getchar();

    srand(time(NULL));
    randomNumber = (rand()%11)-5;
    return randomNumber;

}

And these are the warnings:

warning C4013: 'srand' undefined; assuming extern returning int
warning C4013: 'rand' undefined; assuming extern returning int

Why are these warnings showing up?I'm afraid they might cause some problem later so I would like to eliminate those warnings.

Edits:

Added another library:

#include <stdlib.h>

2 warnings gone, 1 new one:

warning C4244: 'function' : conversion from 'time_t' to 'unsigned int', possible loss of data

New warning solved by change in the srand call to:

srand((unsigned int)time(NULL));

This is the final and solved program:

#include <time.h>
#include<stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[]){

    int randomNumber = 0;
    for (int index = 1; index < argc; index++) {
        printf("The %d is %s\n", index, argv[index]);
    }
    getchar();


    srand((unsigned int)time(NULL));
    randomNumber = (rand()%11)-5;
    return randomNumber;
}

SOLVED,showed the edits step by step for future reference to other users. Thanks

Answer 1

You need to include stdlib.h . That is where those functions are defined.

Answer 2

You're getting this warning because the functions rand and srand haven't been defined.

From the man page:

NAME rand, rand_r, srand - pseudo-random number generator

SYNOPSIS

  #include <stdlib.h> int rand(void); int rand_r(unsigned int *seedp); void srand(unsigned int seed); 

By including stdlib.h, you get the prototypes for these functions, and the compiler can ensure you're calling them correctly.