combine 4 bytes into a 32 bit number - Programmers Heaven

Howdy, Stranger!

It looks like you're new here. If you want to get involved, click one of these buttons!


Welcome to the new platform of Programmer's Heaven! We apologize for the inconvenience caused, if you visited us from a broken link of the previous version. The main reason to move to a new platform is to provide more effective and collaborative experience to you all. Please feel free to experience the new platform and use its exciting features. Contact us for any issue that you need to get clarified. We are more than happy to help you.

combine 4 bytes into a 32 bit number

I have 4 numbers:b1 b2 b3 b4 (b1 is MSB, b4 is LSB) and I want to combine them into one number of 32-bit as b1b2b3b4;tried the following but didn't work (i guess overflow):

res = (unsigned long)((b4 & 0xFF) | (((unsigned long)(b3<<8)) & 0xFF00));
res = (unsigned long)(res | (((unsigned long)(b2<<16)) & 0xFF0000));
res = (unsigned long)(res |(((unsigned long)(b1<<24)) & 0xFF000000));

b1 = 86; b2 = 120; b3 = 154; b4 = 188;
result should be : 1450744508 in decimal or 56789abc in hex


  • TsagTsag Posts: 5Member
    Should work . . .

    I used your code and wrote it in the following way both gave correct outputs for me.


    // 8-bit values
    char a = 0x56;
    char b = 0x78;
    char c = 0x9A;
    char d = 0xBC;

    // combine into a single 32-bit value
    unsigned int value = d | (c<<8) | (b<<16) | (a<<24);
    std::cout << std::hex << value << std::endl;

    output is:


    Compiled in both x64 and x86, result is the same.
    What output are you getting?
  • anoopakesanoopakes Posts: 1Member

    this may not work because you are using signed char.

    change it to unsigned char..

Sign In or Register to comment.