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Is the integer constant's default type signed or unsigned? such as 0x80000000, how can I to decide to use it as a signed integer constant or unsigned integer constant without any suffix?
If it is a signed integer constant, how to explain following case?
printf("0x80000000>>3 : %x\n", 0x80000000>>3); output:
0x80000000>>3 : 10000000 The below case can indicate my platform uses arithmetic bitwise shift, not logic bitwise shift:
int n = 0x80000000; printf("n>>3: %x\n", n>>3); output:
n>>3: f0000000 
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An int is signed by default, meaning it can represent both positive and negative values. An unsigned is an integer that can never be negative.
The C and C++ standards allows the character type char to be signed or unsigned, depending on the platform and compiler. Most systems, including x86 GNU/Linux and Microsoft Windows, use signed char , but those based on PowerPC and ARM processors typically use unsigned char .
An int is always signed in Java, but nothing prevents you from viewing an int simply as 32 bits and interpret those bits as a value between 0 and 264.
The int type in C is a signed integer, which means it can represent both negative and positive numbers. This is in contrast to an unsigned integer (which can be used by declaring a variable unsigned int), which can only represent positive numbers.
C has different rules for decimal, octal and hexadecimal constants.
For decimal, it is the first type the value can fit in: int, long, long long
For hexadecimal, it is the first type the value can fit in: int, unsigned int, long, unsigned long, long long, unsigned long long
For example on a system with 32-bit int and unsigned int: 0x80000000 is unsigned int.
Note that for decimal constants, C90 had different rules (but rules didn't change for hexadecimal constants).
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