C编程：将int *转换为char *

Question

I understand that I cannot convert an int* into a char* but must copy the int* into a char* array using sprintf. My issue is that I believe I am using sprintf correctly but I am getting this warning: warning: passing argument 1 of 'sprintf' from incompatible pointer type. Here is a condensed and simplified version of what I am trying to do.

int* iarray = calloc(top + 1, sizeof(int));
char* carray = (char*)calloc(count,sizeof(char));
/*
some code that adds stuff to the int*, it works fine....something like
for(i=0; i<=count; i++)
    iarray[i] = i;
lets assume that iarray is something like 1234
*/
for(i=0; i<=count; i++)
{
    sprintf(carray, "%d", iarray[i]);
    printf("%s", carray) //this is just to test, it only prints out the first number...
}

This has been driving me nuts... If you want more of my code then please let me know.

Thanks!

EDIT: here is more code. I skipped the basic variable declarations, I can add those in but I assure you all those are fine

Here is a function to do the conversion:

void ints_to_chars(char* carray, int count, int* iarray)
{
    //convert ints to char*
    int i;
    //char *char_array=(char*)calloc(count,sizeof(char));
    printf("ints to chars: char array-->");
    for(i=0; i<=count; i++)
    {
        sprintf(carray, "%d", iarray[i]);
        printf("%s",carray);
    }
    return;
}

int main(int argc, char** argv)
{
int i, j, count;

if(i-1 == 0)
                  bottom = 2;
              else
                  bottom = (int)atoi(argv[i-1])+1;
              top = (int)atoi(argv[i]);
              printf("child %d: bottom=%d, top=%d\n", getpid(), bottom, top);

              prime_iarray = calloc(top + 1, sizeof(int));
              primenumbers(prime_iarray,bottom,top);
              count = prime_iarray[0];

              /*printf("prime int array: \n");
              for(i=1; i<=count; i++)
              {
                  printf("%d", prime_iarray[i]);
              }*/

              //int* into char*
              prime_carray = (char*)calloc(count,sizeof(char));
              ints_to_chars(prime_carray, count, prime_iarray);
              printf("prime char array:\n");
              printf("%s",&prime_carray[0]);

}

EDIT 2: So someone kindly pointed out a silly error that removed the warning I was getting. But my char* only prints out the first character.... I initially was returning a local char* but found out that that will always only return the first character. What am I doing wrong? Can someone explain to me why this is happening? I'm sorry, I'm new to pointers and I'm really struggling with this...

Answer 1

you code is fine. I tested the following which gives the correct result and no compiler warnings,

#include "stdio.h"
#include "stdlib.h"
int main()
{

        int count = 20;
        int top = 10;
        int* iarray = calloc(top + 1, sizeof(int));
        char* carray = (char*)calloc(count,sizeof(char));
        int i;
        int n;

        iarray[0] = 9;
        for(i=0; i<=count; i++)
        {
                        n = sprintf(carray, "%d", iarray[i]);
                        printf("%s", carray); //this is just to test, it only prints out the first number...
        }
}

In your ints_chars function, why do you declare carray to be int* ?

Answer 2

In your example code you have

ints_to_chars(int* carray

carray should be a char* not a int*

Answer 3

You have the wrong signature:

void ints_to_chars(int* carray, int count, int* iarray)
                   ^^ this should be char*

Answer 4

try this :

    char a[100];
    char b[4];
    a[0] = '\0';
    int i;

    for(i=0; i<=count; i++)
    {

            sprintf(b ,"%d" , iarray[i]);
            strcat(a , b);
            strcat(a , "  ");           

    }
printf("%s", a);

by the way, this is an assigment to find prime numbers using multi-process or multi-threaded. ^-^