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Home/ Questions/Q 7840917
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Editorial Team
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Editorial Team
Asked: June 2, 20262026-06-02T15:57:59+00:00 2026-06-02T15:57:59+00:00

I’m studying D language and simultaneously doing comparison to C and C++ languages.It works

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I’m studying D language and simultaneously doing comparison to C and C++ languages.It works fine both dmd and gdc compilers,but when I tested on gcc compiler,I found a thing that looks like bug of GCC compiler,the default initializer of boolean type instead of 0/false see the following code:

C++ code

#include <iostream>
using namespace std;

int main()
{
    bool b;
    cout << b << endl;
    return 0;
}

G++ compiler(gcc version 4.4.3 (Ubuntu 4.4.3-4ubuntu5.1):

g++ -Wall -pedantic test.cpp
test.cpp: In function ‘int main()’: test.cpp:7: warning: ‘b’ is used
uninitialized in this function 
./a.out 64

C code(foo.c):

#include <stdio.h>
#include <stdbool.h>

#define bool _Bool

int main(int argc, char * args[])
{
    bool b;
    printf("%d\n", b);
    return 0;
}

gcc compiler

gcc-4.6 -Wall -pedantic a.c
foo.c: In function ‘main’:
foo.c:9:8: warning: ‘b’ is used uninitialized in this function [-Wuninitialized]
./a.out
64

tcc compiler

tcc -Wall foo.c
./a.out
0

clang compiler

clang -Wall -pedantic foo.c     
./a.out
0

Can someone explain the gcc behavior?

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  1. Editorial Team
    Editorial Team
    2026-06-02T15:58:01+00:00Added an answer on June 2, 2026 at 3:58 pm

    Default initialization in C++ for basic types means “uninitialized”. That is, any value can be there. You got 64 because that just happened to be in that memory location.

    If you want to do value initialization, then you need to use bool():

    bool b = bool(); //Now is false.
    

    Value initialization effectively means initializing basic types to zero.

    C++11 makes this rather cleaner:

    bool b{}; //Now is false.
    
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