Printing a string character-by-character

Question

int main() {
    int i;
    char a[]={"Hello"};
    while(a!='\0') {
        printf("%c",*a);
        a++;
    }
    getch();
    return 0;
}

Strings are stored in contiguous memory locations & while passing the address to printf() it should print the character. I have jst started learning C. I am not able to find an answer to this. Pls help.

The first problem is your title. It's not informative, please edit it. — Maroun, Jul 01 '13 at 13:07
possible duplicate of [What is the difference between char s\[\] and char \*s in C?](http://stackoverflow.com/questions/1704407/what-is-the-difference-between-char-s-and-char-s-in-c) — CB Bailey, Jul 01 '13 at 13:09
Not really an exact duplicate of the question but you are trying to use an array as if it were a pointer so I believe that the answer to the linked question should also answer your question. — CB Bailey, Jul 01 '13 at 13:10
`man putc`, `man fputc` Using `printf` to print a single character is not wrong, but a bit strange. — William Pursell, Jul 01 '13 at 13:29

GeekFactory · Answer 1 · 2013-07-01T13:17:08.697

Well a is the name of the array which you cannot increment. It is illegal to change the address of the array.

So define a pointer to a and then increment

#include <stdio.h>
#include <conio.h>

 int main()
 {
     int i;
     char a[]="Hello";
     char *ptr = a;
     while(*ptr!='\0')
     {
         printf("%c",*ptr);
         // a++ here would be illegal
         ptr++;
     }
     getch();
     return 0;
 }

NOTE:

In fact, arrays in C are non-modifiable lvalues. There are no operations in C that can modify the array itself (only individual elements can be modifiable).

Yu Hao · Answer 2 · 2013-07-01T13:18:49.753

2

In your code, a is a name of an array, you can't modify it like a++. Use a pointer like this:

char *p = "Hello";
while(*p++)
{
     printf("%c",*p);
}

edited Jul 01 '13 at 13:18

answered Jul 01 '13 at 13:09

Yu Hao

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Elazar · Answer 3 · 2013-07-01T13:24:16.003

1

Three problems:

char a[]={"Hello"}; is illegal. {"Hello"} can only initialize char* a[]. You probably want char a[]="Hello";
while(a!='\0') - you probably meant *a != '\0'. a is the array itself.
a++; - an array cannot be incremented. you should increment a pointer pointing to it.

edited Jul 01 '13 at 13:24

answered Jul 01 '13 at 13:17

Elazar

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score 0 · Answer 4 · answered Jul 01 '13 at 14:32

0

You can also try it using a for loop:

#include <stdio.h>

int main(void) {
   char a[] = "Hello";
   char *p;

   for(p = a; *p != '\0'; p++) {
      printf("%c", *p);
   }

   return 0;
}

answered Jul 01 '13 at 14:32

Mestica

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Printing a string character-by-character

4 Answers4