Char array Segmentation fault

Question

I am just playing around with a simple c code. I want to print all the days string in the days array pointer. However I am running to segmentation fault. The problem lies in the for loop. What am I doing wrong here. Thanks

main(){
  const char *days[] = {"Sunday", "Monday", "Tuesday", "Wednesday",\
                        "Thursday", "Friday", "Saturday", "Sunday"};
  int i;

  for(i = 0; i < sizeof(days); ++i)
   printf( "%s\n", days[i]);
}

the `sizeof` operator returns a size in bytes, divide by `sizeof(*days)` to get the number of elements in an array — BeyelerStudios, Apr 08 '16 at 13:55

score 2 · Answer 1 · answered Apr 08 '16 at 13:58

2

int main(){
  const char *days[] = {"Sunday", "Monday", "Tuesday", "Wednesday",\
                        "Thursday", "Friday", "Saturday", "Sunday"};
  size_t i;
  for(i = 0; i < sizeof days / sizeof days[0]; ++i)
   printf( "%s\n", days[i]);
}

sizeof day yield the size of the whole array, so you have to divide it with the size of one of its elements to get the number of element it contains.

answered Apr 08 '16 at 13:58

nalzok

14,965
21
72
139

thanks to all for help – Sunil Gautam Apr 08 '16 at 14:01

score 2 · Answer 2 · answered Apr 08 '16 at 13:58

change the for statement to

for( i = 0; i < ( sizeof(days)/sizeof(days[0]) ); ++i )

The sizeof expression gives you the number of elements in the array, as opposed to the number of bytes it takes

A classic way to do this is a macro

#define SIZEOF( X ) ( sizeof( X )/sizeof( X[0]) )

and then use SIZEOF instead of sizeof

Char array Segmentation fault

2 Answers2