Incrementing integer in a function in C

Question

I have this very simple program

int test(int asdf){
  asdf = asdf + 1;
  return 0;
}

int main ( int argc, char **argv ){

  int a = 1;
  test(a);
  printf("a is %d\n", a);
}

I want the program to output 2, but instead it outputs 1. What went wrong? How should I pass the reference?

@RaymondChen: Beware, the accepted answer to that question is nonsense. — Keith Thompson, Oct 19 '14 at 01:33
@Keith: No, it's quite good, for solving a very different problem (how do I share an object with my callback function?). It simply isn't a duplicate. — Ben Voigt, Oct 19 '14 at 01:33
Rats. I got it from [this question](http://stackoverflow.com/questions/12699712/why-doesnt-return-modify-the-value-of-a-parameter-to-a-function) which is a closer match to the question. — Raymond Chen, Oct 19 '14 at 01:36
That's ok vote to close (because it has been answered already) but why all these downvotes rather just point to duplicate topic? — Jack, Oct 19 '14 at 02:12
@Raymond: Yes, that one is a good duplicate. It doesn't have all the type erasure baggage, just gets to the point of transferring a result from function to caller. — Ben Voigt, Oct 19 '14 at 02:42
@Jack I didn't downvote, but my guess is that the question has many duplicates, so the OP did not do sufficient research before asking. — Raymond Chen, Oct 19 '14 at 02:49

score 1 · Answer 1 · answered Oct 19 '14 at 01:30

Whatever you pass is going to get copied... so you should pass something, that even when copied, lets you find the original a.

That something is a pointer, and its value is the address of a.

If you write your friend's address on a piece of paper, then no matter how many times that paper gets copied, the person holding it can still find the original house.

You write a pointer as int* asdf, and the address of a is written as &a. To find the object when you have a pointer, you need *asdf.

score 1 · Accepted Answer · answered Oct 19 '14 at 01:30

1

First get started with a Pointers chapter from good C book

int test(int* asdf){   // Argument will be pointer to integer
  *asdf = *asdf + 1;   // de-reference and access asdf's content & increment
  return 0;
}

test(&a); // Pass address of a

answered Oct 19 '14 at 01:30

P0W

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This doesn't help a student, because it doesn't even try to explain the concepts. – Ben Voigt Oct 19 '14 at 01:31

score 0 · Answer 3 · edited May 23 '17 at 11:57

0

C doesn't support passing a variable by reference so you can't do that.

However, if you did pass a pointer to a variable then you could get the variable incremented.

Look at:

Passing by reference in C

edited May 23 '17 at 11:57

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answered Oct 19 '14 at 01:31

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Incrementing integer in a function in C

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