char pointer pointing to a char array

Question

#include<stdio.h>
int main(){
 char a[3];
 char *b=NULL;
 a[0]=0;
 a[1]=1;
 a[2]=2;
 b = a;
 printf("%c",b);
 b++;
 printf("%c",b);
 b++;
 printf("%c",b);
 return 0;
}

I tried to print the values 0,1,2 by incrementing the pointer by 1. please help

Sounds like homework. What's your analysis of the problem so far? — Zak
– Zak, Commented Jul 20, 2012 at 14:55

user529758 · Accepted Answer · 2012-07-20 14:55:42Z

4

b is a pointer in itself, you have to dereference it to get the actual values:

printf("%d", *b);
b++;
printf("%d", *b);
b++;

etc.

answered Jul 20, 2012 at 14:55

user529758

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Comments

Oliver Charlesworth · Accepted Answer · 2012-07-20 14:55:23Z

3

%c tells printf to interpret the char argument as a character code (most likely ASCII). Use %d instead.

answered Jul 20, 2012 at 14:55

Oliver Charlesworth

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Comments

BLUEPIXY · Accepted Answer · 2012-07-20 14:57:30Z

2

#include<stdio.h>
int main(){
 char a[3];
 char *b=NULL;
 a[0]='0';
 a[1]='1';
 a[2]='2';
 b = a;
 printf("%c",*b);
 b++;
 printf("%c",*b);
 b++;
 printf("%c",*b);
 return 0;
}

answered Jul 20, 2012 at 14:57

BLUEPIXY

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