gcc strange behaviour on abs function

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I ported some code from windows (vs2010) to gcc

a piece of it looks like:

double r1 /* = some value */; double r2 /* = some value */; double diff = abs(r1-r2); std::cerr<<  r1 << ", " << r2 << ", " << diff<< std::endl; 

it compiles on gcc (arm-linux-gnueabihf-g++ (Raspbian 6.3.0-18+rpi1+deb9u1) 6.3.0 20170516) with -Wall -Wextra without warnings.

The result is:

0.121, 0.0709839, 0 0.015, 0.131958, 0 0.015, 0.00799561, 0 0.21, 0.00799561, 0 0.182, 0.205994, 0 0.015, 0.00799561, 0 

On windows the result is correct. There are double-overloads on the abs functions.

I am not using namespace std;. It seems that under gcc these overloads does not exist in global namespace.

I do not know what exactly the standard says, but i would have expected at least a warning about the double->int-conversion on passing the the difference to the abs function.

Do i have something missed? Why do i not get this warning?

 


abs is a C function that takes one integer and returns another integer.

You want to use the C++ version std::abs instead. Clang would actually warn about the mistake (I even turn it into an error -Werror=absolute-value), not sure if there is a similar flag for gcc.

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