CODEWITHSUNDEEP
cpointers
Ant's
Ant's

Reputation: 13811

Are there any difference between s++ and *s++?

I'm a beginner in C and Let's say I have a code like this:

#include <stdio.h>
void test(char *t)
{
     t++;
    *t = 'e';
}

void main()
{
    char a[] = "anto";
    printf("%c\n",a[1]);
    test(a);
    printf("%c\n",a[1]);
}

This is the sample code, where I am figuring out how pointers work. According to me the statement:

t++;

in the above code will increment the address of array a by 1 char in the calling function test. Fine, now as far I know the * is used to retrieve the object value that the pointer points to.

But weirdly when I change the t++ to 

*t++;

I'm getting the same output as before. I'm literally confused with this, the above statement *t++; should change the contents only know, according to the definition of * operator.

But again this changes the address of t. How come? Where I'm getting the concept wrong?

Thanks in advance

Upvotes: 0

Views: 1500

Answers (4)

John Bode
John Bode

Reputation: 123458

Postfix ++ has higher precedence than unary *, so *t++ is parsed as *(t++); you're dereferencing the result of the expression t++; as a side effect, t is advanced.

Unary * and unary ++ have the same precedence, so if they appear in the same expression they would be evaluated left-to-right. The expression *++t would be parsed as *(++t); you dereference the result of the expression ++t, and as a side effect t is advanced.

++*t would be parsed as ++(*t); you're incrementing the result of the expression *t.

Upvotes: 1

user1199657
user1199657

Reputation:

++ has greater precedence than * so your expression evaluates to *(t++).

Upvotes: 2

tafa
tafa

Reputation: 7326

(*) operator is used to dereference a pointer.

t++;

as a statement returns the previous position of the pointer where as

*t++;

returns the value that the t pointer is pointed to before incrementing it.

The side effects of both statements are same, so you are getting the same output.

*t++;

statement does not change the value t is pointing to because ++ operator has greater precedence then * operator.

Upvotes: 4

Ernest Friedman-Hill
Ernest Friedman-Hill

Reputation: 81684

The expression *t++ is parsed as *(t++) -- the ++ still applies to the pointer, not the contents. The value of t++ is the value of the pointer itself before the increment, while the value of *t++ is what the pointer points to before the increment.

Upvotes: 10

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