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Editorial Team
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Asked: 2026-06-04T20:30:13+00:00

usigned int x=1; signed int y = -1; double z = y * x

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usigned int x=1;
signed int y = -1;
double z = y * x * 0.25;

I’m using Microsoft Visual Studio 10 C++ compiler.
Why z don’t have -0.25 value?
As I saw from disassembly, it makes an signed int multiply (imul), places the result from edx on the stack, and extends it with 0!, as it would be an unsigned int. After that it multiplies it using FP instructions.

.............
imul        edx,dword ptr [ecx]  
mov         dword ptr [ebp-98h],edx  
mov         dword ptr [ebp-94h],0
fild        dword ptr [ebp-98h]  
fmul        qword ptr [__real@3fd0000000000000 (1402FB8h)]  
fstp        qword ptr [z]

Why the result of multiply of signed * unsigned is interpreted as unsigned?

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1 Answer

  1. Editorial Team

    The expression y * x * 0.25 associates as (y * x) * 0.25 in both C and C++.

    When multiplying an unsigned int and a signed int, both operands are converted to unsigned int and the results is also an unsigned int due to the integer conversion rules for arithmetic expressions in both C and C++ so the result of y * x will be UINT_MAX - 1 in both C and C++.

    Your example doesn’t exhibit a compiler bug whether you are compiling your example as C or C++.

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