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在下面的代码中，我使用了 size_t 作为函数参数并传递了负值。我使用以下命令在 GCC(Linux) 上编译了程序。
g++ -Wall size.cpp -o size
GCC 编译成功，没有警告，但结果不是我期望的:
size_t : 18446744073709551615
int : -1
代码:
#include <iostream>
void func1(size_t i)
{
std::cout << "size_t : " << i << std::endl;
}
void func2(int i)
{
std::cout << "int : " << i << std::endl;
}
int main()
{
func1(-1);
func2(-1);
return 0;
}
为什么编译器不为 size_t 的负值生成警告？
最佳答案
因为 size_t 在 C++ 中总是无符号的:
The type
size_tis an implementation-defined unsigned integer type that is large enough to contain the size in bytes of any object.
Why doesn't compiler generate warning for negative value with size_t?
因为为 size_t 分配负值会调用有符号到无符号的转换，这是明确定义的:
If the destination type is unsigned, the resulting value is the least unsigned integer congruent to the source integer (modulo 2n where n is the number of bits used to represent the unsigned type). [Note: In a two’s complement representation, this conversion is conceptual and there is no change in the bit pattern (if there is no truncation). —end note]
关于c++ - 为什么编译器不为 size_t 的负值生成警告？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/46968502/