std::cout打印空的char指针会出现未定义行为

王燕龙(大卫)2024-11-17 13:10

使用std::cout打印char类型的指针，如果这个指针为空的话，那么会导致后边使用std::cout进行的打印都不显示。

如下代码，分别打印非空的char指针，空的char指针，空的int指针，运行结果如下：

（1）打印非空的char指针，执行正常。

（2）打印空的char指针，会导致后边的std::cout打印都打印不出来。

（3）打印空的int指针，没有影响，将空指针当成0打印了出来。

root@wangyanlong-virtual-machine:/home/wangyanlong/cpp# ./a.out 1

1 before print

p=hello

1 after print

root@wangyanlong-virtual-machine:/home/wangyanlong/cpp# ./a.out 2

2 before print

p=root@wangyanlong-virtual-machine:/home/wangyanlong/cpp# ./a.out 3

3 before print

p=0

3 after print

root@wangyanlong-virtual-machine:/home/wangyanlong/cpp#

cpp 复制代码

#include <iostream>
#include <string>
#include <unistd.h>

int main(int32_t argc, char** argv) {
  if(argc != 2) {
    std::cout << "usage:./a.out 1/2/3" << std::endl;
    return 0;
  }

  if (std::string(argv[1]) == "1") {
    char *p = "hello";
    std::cout << "1 before print" << std::endl;
    std::cout << "p=" << p << std::endl;
    std::cout << "1 after print" << std::endl;
  } else if (std::string(argv[1]) == "2") {
    char *p = nullptr;
    std::cout << "2 before print" << std::endl;
    std::cout << "p=" << p << std::endl;
    std::cout << "2 after print" << std::endl;
  } else if (std::string(argv[1]) == "3") {
    int *p = nullptr;
    std::cout << "3 before print" << std::endl;
    std::cout << "p=" << p <<std::endl;
    std::cout << "3 after print" << std::endl;
  }

  return 0;
}