当 %x 读取有符号字符时会发生什么？

Question

i thought *(p3 + 3) will print 90 but it shows ffffff90 why does it happend? i guess MSB is 1, and %x is for reading unsinged hexadecimal integer so it reads 90 like minus integer but it is not clear and i cant find about this problem at printf reference https://cplusplus.com/reference/cstdio/printf/ is there anyone who explain this?

Answer 1

Use an unsigned char * .

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In your environment,

• char is signed.
• char is 8 bits.
• Signed numbers use two's complement.

So you have a char with a bit pattern of 90 ₁₆ . In this environment, that's -112. So you are effectively doing the following:

printf( "%x", (char)-112 );

When passing to variadric function like printf , the smaller integer types are implicitly promoted to int or unsigned int . So what's really happening is this:

printf( "%x", (int)(char)-112 );

So you're passing the int value -112. On your machine, that has the bit pattern FF FF FF 90 ₁₆ (in some unknown byte order). %x expects an unsigned integer, and thus prints that bit pattern as-is.