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Home/ Questions/Q 6946379
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T13:33:28+00:00 2026-05-27T13:33:28+00:00

I’m using an MPC56XX (embedded systems) with a compiler for which an int and

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I’m using an MPC56XX (embedded systems) with a compiler for which an int and a long are both 32 bits wide.

In a required software package we had the following definitions for 32-bit wide types:

typedef   signed int sint32;
typedef unsigned int uint32;

In a new release this was changed without much documentation to:

typedef   signed long sint32;
typedef unsigned long uint32;

I can see why this would be a good thing: Integers have a conversion rank between short and long, so theoretically extra conversions can apply when using the first set of definitions.

My question: Given the above change forced upon us by the package authors, is there a situation imaginable where such a change would change the compiled code, correctly leading to a different result?

I’m familiar with the “usual unary conversions” and the “usual binary conversions”, but I have a hard time coming up with a concrete situation where this could really ruin my existing code. But is it really irrelevant?

I’m currently working in a pure C environment, using C89/C94, but I’d be interested in both C and C++ issues.

EDIT: I know that mixing int with sint32 may produce different results when it’s redefined. But we’re not allowed to use the original C types directly, only the typedef’ed ones.
I’m looking for a sample (expression or snippet) using constants, unary/binary operators, casts, etc. with a different but correct compilation result based on the changed type definition.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-27T13:33:28+00:00Added an answer on May 27, 2026 at 1:33 pm

    It might lead to subtle issues because literal numbers are int by default.

    Consider the following program:

    #include <iostream>
    
    typedef signed short old16;
    typedef signed int old32;
    
    void old(old16) { std::cout << "16\n"; }
    void old(old32) { std::cout << "32\n"; }
    
    typedef signed short new16;
    typedef signed long new32;
    
    void newp(new16) { std::cout << "16\n"; }
    void newp(new32) { std::cout << "32\n"; }
    
    int main() {
      old(3);
      newp(3); // expected-error{{call of overload ‘newp(int)’ is ambiguous}}
    }
    

    This leads to an error because the call to newp is now ambiguous:

    prog.cpp: In function ‘int main()’:
    prog.cpp:17: error: call of overloaded ‘newp(int)’ is ambiguous
    prog.cpp:12: note: candidates are: void newp(new16)
    prog.cpp:13: note:                 void newp(new32)
    

    whereas it worked fine before.

    So there might be some overloads surprises where literals were used. If you always use named (and thus typed) constants, you should be fine.

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