CODEWITHSUNDEEP
c
stickfigure4
stickfigure4

Reputation: 185

Dereferencing pointers confusion

I'm trying to learn about pointers in the C language. But I have a question regarding dereferencing them.

This piece of code works fine, it displays 'i' as I desire.

Code-1

#include <stdio.h>

void myfunction(char** str);

int main()
{
    char *name = "Gianni"; 
    myfuntion(&name); 
    return 0;
}

void myfunction(char** str){

    char* test = *str;
    printf("%c", test[1]);
}

But what I guessed to have the same result turns out to be a little bit different. My question is why do I not get the same result from the second piece of code?

I dereferenced the pointer so it points to the first character again and can read it until the '\0' terminator. So I thought it would be possible to display the single characters here as well.

Code-2

#include <stdio.h>

void myfunction(char** str);

int main()
{
    char *name = "Gianni"; 
    myfuntion(&name); 
    return 0;
}

void myfunction(char** str){
    printf("%c", *str[1]);
}

Upvotes: 1

Views: 78

Answers (1)

user2736738
user2736738

Reputation: 30906

This is precedence issue.(Array subscript [] has higher precedence than dereference *). You should try 

(*str)[1]

This will give you the correct result. In earlier case (In Code-2) you had undefined behavior. You were de-referencing str[1] which is pointing to out of bound memory.

So initially what you were doing in Code-2

*str[1] --> *(*(str+1))

Now what we did,

(*str)[1] --> *((*str)+1)

Well in the modification you are basically getting the address of the array (string literals are array which decayed into pointer) and then you index into correct position - which will give you the correct result.

Well if you check earlier example you will see the similarity. Earlier code (Code-1) you print temp[1] which is *(temp+1) But what is temp? In the line before you did this temp = *str so what is it you are doing?

*(temp+1) --> *(*str+1) --> *((*str)+1)

Upvotes: 5

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