Reporting error on calling memcpy

Question

How can I make sure that if someone uses come specific function(say memcpy) in code, then it returns an error.

We have removed all the instances of memcpy from code with some internal designed function,

what I want to make sure is that whenever someone uses memcpy in future, the compiler throws an errors msg.

You have two good answers, but I'm curious why you want to ban memcpy? It's a pretty simple function and should be hard to get wrong. — nonsensickle, Jul 11 '13 at 11:11
if you use GCC, then add `#pragma GCC poison memcpy` in you main head file, it may help you — How Chen, Jul 11 '13 at 11:13
to overcome the problem of overlap, yes we can use memmove but they dont use it...i dont know why — Sohrab Ahmad, Jul 11 '13 at 11:26

score 5 · Accepted Answer · answered Jul 11 '13 at 11:06

5

You can use the preprocessor for this, like

#define memcpy(a, b, c) do_not_use_memcpy

Put that in a header file that is included in all source files, and the preprocessor will replace all calls to memcpy with the (undefined) symbol do_not_use_memcpy. As that symbol is undefined, you will get a compiler error about it.

answered Jul 11 '13 at 11:06

Some programmer dude

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The problem with this approach is that the newly created function uses memcpy internally. – Sohrab Ahmad Jul 11 '13 at 11:09
+1 particularly nice in that it will give you a readable compiler warning containing the text "do_not_use_memcpy" – Bathsheba Jul 11 '13 at 11:09
@Sohrab Ahmad, undef and re-def it in the new function. – Bathsheba Jul 11 '13 at 11:10
@SohrabAhmad For that single place, undefine the macro with `#undef memcpy`. – Some programmer dude Jul 11 '13 at 11:10
It's sound dangerous to me because gcc may replace some loops to a call to the memcpy function. This is done through the tree-loop-distribute-patterns optimization in gcc 4.8. – Maxime Chéramy Jul 11 '13 at 11:30
1

@Maxime The code optimizations are done long after the preprocessor has run, so that won't be a problem. This doesn't actually replace the actual `memcpy` function, just changes the calls to it. – Some programmer dude Jul 11 '13 at 11:32
I see, but are you sure there won't be any call to the memcpy at the end? What about adding a `-fno-builtin-memcpy`? – Maxime Chéramy Jul 11 '13 at 11:41
1

@Maxime If that macro is in all translation units, then there won't be any explicit calls to `memcpy`. That flag just tells GCC that it should not use its own implementation of `memcpy`, and instead use the one in the standard library. – Some programmer dude Jul 11 '13 at 11:45

score 4 · Answer 2 · answered Jul 11 '13 at 11:06

4

To avoid breaking libraries, use the deprecated attribute :

void * my_new_memcpy ( void * destination, const void * source, size_t num )
{
    return memcpy(destination, source, num);
} __attribute__((deprecated));

// Make sure this is used *after* declaring the function
#define memcpy my_new_memcpy

answered Jul 11 '13 at 11:06

Tom van der Woerdt

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What about using "-Wl,-wrap,memcpy" instead? – Maxime Chéramy Jul 11 '13 at 11:21
1

@Maxime Not sure: that's a linker argument while the `__attribute__((deprecated))` is (iirc) interpreted by the compiler. – Tom van der Woerdt Jul 11 '13 at 11:27
I meant, instead of using the `#define` but still using the deprecated attribute (I don't know if possible though). – Maxime Chéramy Jul 11 '13 at 11:37
1

@Maxime I knew you meant that :-) But you can't, because `deprecated` is interpreted in the compilation phase which goes before the linking phase (which is where your argument runs). – Tom van der Woerdt Jul 11 '13 at 11:39

Reporting error on calling memcpy

2 Answers2