How to print a char array in C through printf?

Question

This results in segmentation fault. What needs to be corrected?

int main(void)
{
    char a_static = {'q', 'w', 'e', 'r'};
    char b_static = {'a', 's', 'd', 'f'};

    printf("\n value of a_static: %s", a_static);
    printf("\n value of b_static: %s\n", b_static);
}

Answer 1

The code posted is incorrect: a_static and b_static should be defined as arrays.

There are two ways to correct the code:

• you can add null terminators to make these arrays proper C strings:

  #include <stdio.h>
  
  int main(void) {
      char a_static[] = { 'q', 'w', 'e', 'r', '\0' };
      char b_static[] = { 'a', 's', 'd', 'f', '\0' };
  
      printf("value of a_static: %s\n", a_static);
      printf("value of b_static: %s\n", b_static);
      return 0;
  }
  

• Alternately, printf can print the contents of an array that is not null terminated using the precision field:

  #include <stdio.h>
  
  int main(void) {
      char a_static[] = { 'q', 'w', 'e', 'r' };
      char b_static[] = { 'a', 's', 'd', 'f' };
  
      printf("value of a_static: %.4s\n", a_static);
      printf("value of b_static: %.*s\n", (int)sizeof(b_static), b_static);
      return 0;
  }
  

  The precision given after the . specifies the maximum number of characters to output from the string. It can be given as a decimal number or as * and provided as an int argument before the char pointer.

Answer 2

This results in segmentation fault. ? because of the below statement

char a_static = {'q', 'w', 'e', 'r'};

a_static should be char array to hold multiple characters. make it like

 char a_static[] = {'q', 'w', 'e', 'r','\0'}; /* add null terminator at end of array */

Similarly for b_static

char b_static[] = {'a', 's', 'd', 'f','\0'};

Answer 3

You need to use array instead of declaring

a_static
b_static

as variables

So it look like this:

int main()
{
  char a_static[] = {'q', 'w', 'e', 'r','\0'};
  char b_static[] = {'a', 's', 'd', 'f','\0'};
  printf("a_static=%s,b_static=%s",a_static,b_static);
  return 0;
}

Answer 4

The thing is that you are using C Style Strings, and a C Style String is terminated by a zero. For example if you'd want to print "alien" by using a char array:

char mystring[6] = { 'a' , 'l', 'i', 'e' , 'n', 0}; //see the last zero? That is what you are missing (that's why C Style String are also named null terminated strings, because they need that zero)
printf("mystring is \"%s\"",mystring);

The output should be:

mystring is "alien"

Back to your code, it should look like:

int main(void) 
{
  char a_static[5] = {'q', 'w', 'e', 'r', 0};
  char b_static[5] = {'a', 's', 'd', 'f', 0}; 
  printf("\n value of a_static: %s", a_static); 
  printf("\n value of b_static: %s\n", b_static); 
  return 0;//return 0 means the program executed correctly
}

By the way, instead of arrays you can use pointers (if you don't need to modify the string):

char *string = "my string"; //note: "my string" is a string literal

Also you can initialize your char arrays with string literals too:

char mystring[6] = "alien"; //the zero is put by the compiler at the end 

Also: Functions that operate on C Style Strings (e.g. printf, sscanf, strcmp,, strcpy, etc) need zero to know where the string ends

Hope that you learned something from this answer.