Understanding memory addresses and pointers in C

Question

I received the following question on one of my practice problemswhere it says to determine what is printed by this code:

#include <stdio.h>
int main(){
    int ids[3] = {100,200,300};
    int *salary, salary1, salary2, *salary3;
    salary1 = ids[0] * ids[1];
    salary = &ids[1] ;
    salary2 = *(ids+1)* *(ids+2);
    salary3 = ids+2;
    printf("*salary = %d\nsalary1 = %d\n", *salary, salary1);
    printf("salary2 = %d\nsalary3 = %p\n", salary2, salary3);
}

I am a bit confused about this code. Firstly, in line 4:

    int *salary, salary1, salary2, *salary3;

Why is there an asterisk at the beginning of salary3 if an asterisk was already used in the beginning of the line?

Secondly when it says:

    salary1 = ids[0] * ids[1];

how are we supposed to determine the value of salary1 when we don't know the value of ids[1]?

The * only applies to the variable right after it, not all the variables on the line. — Barmar
– Barmar, Commented Feb 15, 2017 at 23:14
Because it is not int* salary etc, but int *salary etc. The * qualifies the variable, not the type. — Weather Vane
– Weather Vane, Commented Feb 15, 2017 at 23:14

Stack Exchange Broke The Law · Accepted Answer · 2017-02-15 23:15:22Z

2

why is there an asterisk at the beginning of salary3 if an asterisk was already used in the beginning of the line?

The asterisk means the next variable is a pointer. It's the same as:

int *salary;
int salary1;
int salary2;
int *salary3;

but on one line.

how are we supposed to determine the value of salary1 when we don't know the value of ids[1]?

But you do know the value of ids[1]. It's 200.

answered Feb 15, 2017 at 23:15

Stack Exchange Broke The Law

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3 Comments

Shoaib Ahmed Over a year ago

I know the value of ids[0] which is 2000, but not ids[1]

Weather Vane Over a year ago

@ShoaibAhmed no, ids[0] is 100. Look at the array int ids[3] = {100,200,300}; which defines ids[0] as 100, ids[1] as 200, ids[2] as 300.

Shoaib Ahmed Over a year ago

oh my bad, i miss read that part. Thank you for pointing that out.

Malcolm McLean · Accepted Answer · 2017-02-16 00:41:44Z

1

Firstly, C declaration syntax is a bit confusing

int *x, y;

Means x is a pointer to an int, y is an int. That's such a source of confusion that virtually every style guide forbids it.

This

  int ids[3] = {100,200,300};

declares an array. ids[0] = 100, ids[1] = 200, ids[2] = 300. It's only really possible to match subscripts to array contents by eye when arrays are small. So this isn't a normal use of arrays. Either you have large read-only tables in global memory, or you create an array at runtime and fill it. However there shouldn't be an issue is ids[1] holding 200, that's how C array syntax works.

answered Feb 16, 2017 at 0:41

Malcolm McLean

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