如何使用std :: cout模拟printf的％p格式？ [英] How to simulate printf's %p format when using std::cout?

问题描述

unsigned char *teta = ....;
...
printf("data at %p\n", teta); // prints 0xXXXXXXXX

如何使用 iostream s？是否有 std :: ？特征像 std :: hex 来做这种转换（地址 - >字符串），所以 std :: cout< std :: ??? << teta < std :: endl 会打印该地址吗？

How can I print variable address using iostreams? Is there a std::??? feature like std::hex to do this kind of conversion (address -> string), so std::cout << std::??? << teta << std::endl will print that address?

（没有sprintf的，请;））

(no sprintf's, please ;))

推荐答案

投入 void * ：

unsigned char* teta = ....;
std::cout << "data at " << static_cast<void*>(teta) << "\n";

iostreams通常假设你有一个字符串 char * 指针，但是 void * 指针只是一个地址（简化），所以iostreams不能做任何事情，除了将该地址转换为字符串，而不是该地址的内容。

iostreams generally assume you have a string with any char* pointer, but a void* pointer is just that - an address (simplified), so the iostreams can't do anything other than transforming that address into a string, and not the content of that address.

这篇关于如何使用std :: cout模拟printf的％p格式？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！