相关疑难解决方法(0)

假设这些代码编译成g++:

#include <stdlib.h>

int main() {
    int a =0;

    goto exit;

    int *b = NULL;

exit:
    return 0;
}

g++ 会抛出错误:

goto_test.c:10:1: error: jump to label ‘exit’ [-fpermissive]
goto_test.c:6:10: error:   from here [-fpermissive]
goto_test.c:8:10: error:   crosses initialization of ‘int* b’

似乎goto无法跨越指针定义,但gcc编译好,没有任何抱怨.

修复错误之后,我们必须在任何goto语句之前声明所有指针,也就是说你必须声明这些指针,即使你现在不需要它们(并且违反了一些原则).

什么原因设计考虑g++禁止有用的尾部goto声明？

更新:

goto可以交叉变量(任何类型的变量,不限于指针)声明,但除了那些获得初始化值.如果我们删除NULL上面的任务,g++请立即保持沉默.因此,如果要声明goto-cross-area 之间的变量,请不要初始化它们(并且仍然违反某些原则).

我最大的问题如上所述，无法跳转到fin标签（第27行错误），错误：从此处（第12和14行错误）和交叉初始化错误（第20行错误），请帮忙！

#include <iostream>
#include <string>

int main()
{
  std::string name;
  std::cout << "Please comply. y/n: ";
  std::string answer;
  std::cin >> answer;
  if (answer == "y"){std::cout << "You were spared." << std::endl; goto fin;}
  if (answer == "Miche"){std::cout << "The killers understood that you understood the prophecy, so they took you to their master" << std::endl; goto secret;}
  if (answer == "n"){std::cout << "You were brutally killed." << std::endl; goto fin;}
  else {std::cout << "You randomly babled " << answer << …