Question

I can’t assign values to a char array in loop, doing it contains same value in every array value

for example this works

char* foo[3];

foo[0] = "mango"; foo[1] = "kiwi"; foo[2] = "banana";

int i=0; for(i=0;i<3;i++) 
{
   printf("%s\n",foo[i]); 
}

but this doesn’t and I don’t understand why.

char* foo[3]; int i=0;

for(i=0;i<3;i++) {
  char temp[5];
  sprintf(temp,"VAL:%d",i);
  foo[i] = temp; 
}

for(i=0;i<3;i++) 
{
  printf("%s\n",foo[i]); 
}

please help and thanks in advance

Answer 1

You have to remember that in C, a char* does not actually store a string, but it stores the address of a memory location that will be treated as the first character of a string.

In your first example, each element of the foo array holds the address of a different string-literal.
In your second example, each element of the foo array is made to point to the local variable temp. Although each iteration through the loop results in a separate instance of temp, any sane compiler will place all those instances on top of each other, thus giving the results you experience.

The solution is either to use a 2D array:

char foo[3][6]; int i=0;

for (i=0; i<3; i++) { sprintf(foo[i],"VAL:%d", i); }

for (i=0; i<3; i++) { printf("%s\n", foo[i]); }

Or to use dynamic allocation:

char* foo[3]; int i=0;

for (i=0; i<3; i++) { char* temp = malloc(6); sprintf(temp, "VAL:%d", i); foo[i] = temp; }

for (i=0; i<3; i++) { printf("%s\n", foo[i]); free(foo[i]); }