CODEWITHSUNDEEP
c
karthik A
karthik A

Reputation: 655

C String prints junk characters

#include <stdio.h>
#include <stdlib.h>
int main()
{
    char a[]="shevchenko ac milan";
printf("%s",&a);
}

This is prints "shevchenko ac milan"

but 
#include <stdio.h>
#include <stdlib.h>
int main()
{
    char a[]="shevchenko ac milan";
printf("%s",&a+1);
}

Why does this print junk characters?

Upvotes: 2

Views: 3412

Answers (4)

Algorithmist
Algorithmist

Reputation: 6695

This is happening because when you say &a + 1 .This would make the pointer reach one location ahead of the end of string.This is the pointers basic operation.You could check this for an integer array also.

For e.g.

int a[]={1,2,3,4,5};

printf("%d",&a+1); would always print garbage value.

That is it would now going to point to next 1D array which is actually not there.Thus garbage value gets printed.

Upvotes: 1

BMitch
BMitch

Reputation: 263469

When you increment the "address that points to an address", you get an unassigned value. fbrereto is right, you only need to pass a for the pointer to the first position, and a+1 to point to the second position in the array.

Upvotes: 0

peoro
peoro

Reputation: 26060

&a is of type pointer to a char[20].

When you do &a+1 you'll go to the next char[20] item in memory, thus you'll go after a.

You should instead have a char*: by summing 1 to it you'll go to the next char.

To obtain a char* you can just use a (it decays in a char* by doing this), and thus a+1 to go the next char.

Upvotes: 10

fbrereto
fbrereto

Reputation: 35925

You don't need to pass the address of a (&a) - in C a string is an array of chars, so a is already an address. Try:

char a[]="shevchenko ac milan"; printf("%s", a);

and

char a[]="shevchenko ac milan"; printf("%s", a+1);

Upvotes: 3

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