在C中声明大小数组

Question

as much as i know the next code should not run.

The first problem is declaring 'int i' inside the loop and not at the start code.

The next problem is the array size is defined in run time and therefor we must use dynamiclly allocation.

I was expected to get "segment fualt".

I run it with the command:

gcc -Wall commandLineArgument.c -o ex1

int main(void){
  int size=0;
  printf("enter number:\n");
  scanf("%d",&size);
  printf("The size is: %d\n",size);
  int arr[size];
  for(int i= 0;i<size;i++)
  {
    arr[i] = 5;
  }

  for(int i= 0;i<size;i++)
  {
    printf("%d,",arr[i]);
  }  
  printf("\n");
  return 0;}

edit if variable length arrays have been supported in C since C99 then how the update code running?

gcc -Wall -std=c89 commandLineArgument.c -o ex1 

int main(void){
int i;
int size=0;
printf("enter number:\n");
scanf("%d",&size);
printf("The size is: %d\n",size);
int arr[size];
printf("The size of arr: %lu\n",sizeof(arr));
for(i= 0;i<size;i++)
{
    arr[i] = 5;
}

for(i= 0;i<size;i++)
{
    printf("%d,",arr[i]);
}    
printf("\n");
return 0;}

Answer 1

This is anachronistic: your code has been valid C since and including C99.

Since at least C99 you are able to declare i locally in the for loop.

And variable length arrays have been supported in C since C99. (Note that this is not supported even in C++17).

Answer 2

It looks like you are using an old version of gcc.

The two features you are trying to use were added in C99. The default for gcc versions older than version 5 is to compile using the old C89 standard. To use features added in later standards you need to explicitly tell older versions of gcc to use a newer standard.

The compiler option -std=c11 instructs gcc to use the current C11 standard. The minimum you need is -std=99 .

I would recommend that you tell gcc to use the current standard:

 gcc -Wall -std=c11 commandLineArgument.c -o ex1

Another way to solve your problem is to upgrade to gcc v5 or newer.