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Home/ Questions/Q 7847739
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Editorial Team
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Editorial Team
Asked: June 2, 20262026-06-02T17:59:29+00:00 2026-06-02T17:59:29+00:00

I’m trying to convert an int to an byte array and then do base64

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I’m trying to convert an int to an byte array and then do base64 to create a blockId for Azure Rest API. I’ve gotten the first bit correct, the when where I convert the int to a base64 string:

int a = 127;
int b = 4000;

C#:

byte[] blockIdBytes = BitConverter.GetBytes(a);
string blockIdBase64 = Convert.ToBase64String(blockIdBytes);

a gives “fwAAAA==” and b gives “oA8AAA==”

C++

QByteArray temp;

for(int i = 0; i < sizeof(a); i++) {
    temp.append((char)(a >> (i * 8)));
}

a gives “fwAAAA==” and b gives “oA8AAA==” (same values as above, which is correct)

Now the issue is, when I try to convert the base64-string back to an int? My bytearray to int method does not work on numbers bigger than 127, why?

int result = 0;

for(int i = temp.size(); i >= 0; i--) {
    result = (result << 8) + temp[i];
}

127 works but when I do 128 (for example) the result is “-128”. I realize the it overflows, but why and where?

EDIT:

Tried:

QByteArray temp;
int a = 340;

for(int i = 0; i < sizeof(a); i++) {
  temp.append((unsigned char)(a >> (i * 8)));
}

Which actaully gives “340” when I convert it back, “255” gives “-1” and “256” gives “256”

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-02T17:59:30+00:00Added an answer on June 2, 2026 at 5:59 pm

    when you are converting back you need to treat all the values in temp[i] as unsigned char
    or ignore the signed bit . In the below code snippet you end up promoting temp[i] to integer then explicitly we reset the signed bit if any to 0 making it positive

    result = (result << 8) + ((int)(temp[i]) & 0xFF)

    you should be able to achieve the same using

    result = (result << 8) + ((unsigned char)(temp[i]))

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