由于在一个非英语国家，我想做一个测试与字符数组和非ASCII字符。
我用MSVC和Mingwin GCC编译了这段代码：
#include <iostream>
int main()
{
constexpr char const* c = "é";
int i = 0;
char const* s;
for (s = c; *s; s++)
{
i++;
}
std::cout << "Size: " << i << std::endl;
std::cout << "Char size: " << sizeof(char) << std::endl;
}两者都显示Char size: 1，但MSVC显示Size: 1，明温GCC显示Size: 2。
这是由非ASCII字符引起的未定义行为，还是有其他原因（可能是UTF-8中的GCC编码和UTF-16中的MSVC）？
1条答案
按热度按时间deikduxw1#
The encoding used to map ordinary string literals to a sequence of code units is (mostly) implementation-defined.
GCC defaults to UTF-8 in which the character
éuses two code units and my guess is that MSVC uses code page 1252 , in which the same character uses up only one code unit. (That encoding uses a single code unit per character anyway.)Compilers typically have switches to change the ordinary literal and execution character set encoding, e.g. for GCC with the
-fexec-charsetoption.Also be careful that the source file is encoded in an encoding that the compiler expects. If the file is UTF-8 encoded but the compiler expects it to be something else, then it is going to interpret the bytes in the file corresponding to the intended character
éas a different (sequence of) characters. That is however independent of the ordinary literal encoding mentioned above. GCC for example has the-finput-charsetoption to explicitly choose the source encoding and defaults to UTF-8.If you intent the literal to be UTF-8 encoded into bytes, then you should use
u8-prefixed literals which are guaranteed to use this encoding:Note that the type
autohere will beconst char*in C17, butconst char8_t*since C20.smust be adjusted accordingly. This will then guarantee an output of2for the length (number of code units). Similarly there areuandUfor UTF-16 and UTF-32 in both of which only one code unit would be used foré, but the size of code units would be 2 or 4 bytes (assumingCHAR_BIT == 8) respectively (typeschar16_tandchar32_t).