CODEWITHSUNDEEP
c
abdullahDev
abdullahDev

Reputation: 39

How can I print each character in my character array using pointers?

I wrote the following code:

char arrayD[] = "asdf";
char *arraypointer = &arrayD;
while(*arraypointer != '\0'){
    printf("%s \n", arraypointer+1);
    arraypointer++;
}

I tried %d %c to print each character. However, with %c I get "? ? ? ?", with %s I get "sdf sd f ". etc. What am I missing here?

Upvotes: 0

Views: 144

Answers (3)

fist ace
fist ace

Reputation: 47

hey you were not dereferncing the pointer. you did arrayptr but you need to do *arrayptr. Also you need to use %c if you are printing single character.

here is the code

#include <stdio.h>
int main()
{
    char arrayD[] = "asdf";
    char *arraypointer = arrayD;
    while(*arraypointer != '\0'){
        printf("%c \n", *(arraypointer));
        arraypointer++;
    }
    return 0;
}

output :

a 
s 
d 
f

Upvotes: 0

0___________
0___________

Reputation: 68013

void print_string_by_chars(const char *str)
{
    while(*str)
    {
        putc(*str++, stdout);
        putc('\n', stdout);
    }
}

usage in your case:

print_string_by_chars(arrayD)

Upvotes: 0

Cristian Bidea
Cristian Bidea

Reputation: 709

You're printing pointer addresses, instead of what the pointer is pointing to. Also arrayD is the address, you don't need &arrayD. Here is a complete working sample:

#include <stdio.h>
int main()
{
    char arrayD[] = "asdf";
    char *arraypointer = arrayD;
    while(*arraypointer != '\0'){
        printf("%c \n", *(arraypointer+1));
        arraypointer++;
    }
    return 0;
}

Upvotes: 2

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