CODEWITHSUNDEEP
c++windowsprintf
Zachary
Zachary

Reputation: 1703

why printf output zero for a non-zero double?

I have a piece of code:
Under Windows MSVC 2012

#include <stdio.h>
#include <string.h>


namespace myname{
    double var = 42;
}

extern "C" double _ZN6myname3varE = 10.0;

int main(){
    printf("%d\n", _ZN6myname3varE);

    return 0;
}

The output is 0. But I think the output should be 10. Could you help explain why?

Upvotes: 1

Views: 417

Answers (2)

Yu Hao
Yu Hao

Reputation: 122383

printf("%d\n", _ZN6myname3varE);

%d should be changed to %f, which prints type double

But there's another problem: name mangling. When I tested the program in gcc, it shows an error:

Error: symbol `_ZN6myname3varE' is already defined

The problem is the name _ZN6myname3varE is a reserved identifier in C++ because it begins with underscore and an uppercase letter.

Upvotes: 0

raj raj
raj raj

Reputation: 1922

Answer for "But I want to know why 0 is output? How this happen?".

double is 64-bit and int is 32-bit. When double is truncated to int (because of using %d), only first 4 bytes stored in the memory location of double is taken into the int value.

Here, value of double _ZN6myname3varE is 10.0, which is 0x4024000000000000 in hex and stored as 00000000 00002440 in memory (little endian). So, when truncated to int, only 4 byte LSB is taken which is obviously zero.

Upvotes: 2

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