integer-promotion

问题 I made a statement to a colleague of mine, which was: "chars are automatically promoted to integers in C expressions, and that's fine for performance since CPUs work fastest with their natural word size. I believe char promotion behavior is stated somewhere in the standard due to a char's rank. This is the response I got back: "Characters are not default promoted to an integer. The register size is 32 bit, but multiple byte values in a row can be packed into a single register as a compiler

问题 I made a statement to a colleague of mine, which was: "chars are automatically promoted to integers in C expressions, and that's fine for performance since CPUs work fastest with their natural word size. I believe char promotion behavior is stated somewhere in the standard due to a char's rank. This is the response I got back: "Characters are not default promoted to an integer. The register size is 32 bit, but multiple byte values in a row can be packed into a single register as a compiler

问题 This question already has answers here : what does it mean to bitwise left shift an unsigned char with 16 (2 answers) Closed 12 months ago . I am trying to understand shift operators in C/C++, but they are giving me a tough time. I have an unsigned 8-bit integer initialized to a value, for the example, say 1. uint8_t x = 1; From my understanding, it is represented in the memory like |0|0|0|0|0||0||0||1| . Now, when I am trying to left shit the variable x by 16 bit, I am hoping to get output 0

问题 This question already has answers here : Implicit type conversion rules in C++ operators (9 answers) Closed 5 years ago . Why is unsigned short * unsigned short converted to int in C++11? The int is too small to handle max values as demonstrated by this line of code. cout << USHRT_MAX * USHRT_MAX << endl; overflows on MinGW 4.9.2 -131071 because (source) USHRT_MAX = 65535 (2^16-1) or greater* INT_MAX = 32767 (2^15-1) or greater* and (2^16-1)*(2^16-1) = ~2^32 . Should I expect any problems

来源： https://stackoverflow.com/questions/63908544/is-it-ub-to-give-a-char-argument-to-printf-where-printf-expects-a-int

来源： https://stackoverflow.com/questions/63908544/is-it-ub-to-give-a-char-argument-to-printf-where-printf-expects-a-int

问题 This question already has an answer here : C++ size_t modulus operation with negative operand (1 answer) Closed 16 days ago . #include <iostream> #include <string> #include <vector> using namespace std; int main() { vector<int> v = {1, 2, 3, 4, 5, 6, 7}; int i = -4; cout << i << endl; cout << v.size() << endl; cout << i % v.size() << endl; cout << -4 % 7 << endl; } The above code prints: -4 7 5 -4 Can someone please explain why i % v.size() prints 5 instead of -4 ? I'm guessing it has

问题 I have the following code: unsigned char x = 255; printf("%x\n", x); // ff unsigned char tmp = x << 7; unsigned char y = tmp >> 7; printf("%x\n", y); // 1 unsigned char z = (x << 7) >> 7; printf("%x\n", z); // ff I would have expected y and z to be the same. But they differ depending on whether a intermediary variable is used. It would be interesting to know why this is the case. 回答1: This little test is actually more subtle than it looks as the behavior is implementation defined: unsigned

问题 Motivated from a code snippet on this blog under "What happens when I mix signed and unsigned integers?" I decided to run it with few different values of signed and unsigned integers and observe the behaviour. Here is the original snippet (slightly modified, however intent is still same) #include <stdio.h> int main(void) { unsigned int a = 6; int b = -20; int c = (a+b > 6); unsigned int d = a+b; printf("<%d,%u>", c, d); } OUTPUT: <1,4294967282> Now when I run the same program for a = 6 and b

问题 I'm using IAR Workbench compiler with MISRA C:2004 checking on. The fragment is: #define UNS_32 unsigned int UNS_32 arg = 3U; UNS_32 converted_arg = (UNS_32) arg; /* Error line --> */ UNS_32 irq_source = (UNS_32)(1U << converted_arg); The MISRA error is: Error[Pm136]: illegal explicit conversion from underlying MISRA type "unsigned char" to "unsigned int" (MISRA C 2004 rule 10.3) I don't see any unsigned char in any of the code above. The discussion at Why did Misra throw an error here?