Why does this segfault in C? -

i can't figure out why tiny c program segfaults:

#include <stdio.h> #include <stdlib.h>  int main(int argc, char *argv[]){      int in = atoi(argv[1]);     printf("input %d\n",in);      int *n = (int *)malloc(in);     int j;      (j=0;j<in;j++)         n[j] = j;      printf("sanity check...\n");      char *c = (char *)malloc(1024*1024*20);     int i;     (i=0; i<20*1024*1024;i++)         c[i] = i;      printf("no segfault. yay!\n");      return 0; } 

compiled with:

$ gcc -o0 test.c -o run

output:

$ ./run 1000

$ input 1000

$ sanity check...

$ [1] 17529 segmentation fault (core dumped) ./run 1000

now if move 1 of for-loops down this:

#include <stdio.h> #include <stdlib.h>  int main(int argc, char *argv[]){      int in = atoi(argv[1]);     printf("input %d\n",in);      int *n = (int *)malloc(in);     int j;      printf("sanity check...\n");      char *c = (char *)malloc(1024*1024*20);     int i;     (i=0; i<20*1024*1024;i++)         c[i] = i;      printf("no segfault. yay!\n");      (j=0;j<in;j++)         n[j] = j;      return 0; } 

everything works.. same compilation step, output:

$ ./run 1000

$ input 1000

$ sanity check...

$ no segfault. yay!

reason why i'm doing large 20mb malloc try , remove cache effects code profiling. feels both implementations should work, first 1 segfaults when malloc-ing 20mb array. missing obvious here?

thanks.

int in = atoi(argv[1]); int *n = (int *)malloc(in); 

you're allocating in bytes, not in integers. try:

malloc(sizeof(int) * in); 

your second allocation works because sizeof(char) 1.