我想从基于子函数的退出值的主函数打印

Question

我想打印由算术计算生成的不同打印语句以及3个不同的错误（错误的格式，除以零和错误的操作符）。

我负责的，现在的问题是，无论输入是什么，程序从出口（50）打印出“错误陈述”，即使格式是正确的......

这里我的代码：

#include <stdio.h> 
#include <unistd.h> 
#include <stdlib.h> 
#include <sys/wait.h> 

int childFunction(char *string); 

char input; 
char *string = &input; 
int status; 
int pid; 

int main (void) 
{ 
    printf("This program makes simple arithmetics\n"); 

    while(1) 
    { 
     printf("Enter an arithmetic statement, e.g., 34 + 132\n"); 
     input = scanf("%s",&input); 

     pid = fork(); 

     if (pid == -1) // Error creating child process 
     { 
      perror("Impossible to fork\n"); 
      exit(0); 
     } 

     else if (pid == 0) // Child process 
      childFunction(&input); 

     else 
     { 
      if (pid > 0) // Parent process 
      { 
       printf("Created a child to make your operation, waiting\n"); 
       wait(&status); 

       sleep(2); 

       if (WEXITSTATUS(status) == 50) 
       printf("Wrong Statement\n\n"); 

       else if (WEXITSTATUS(status) == 100) 
       printf("Division by zero\n\n"); 

       else 
       if (WEXITSTATUS(status) == 200) 
        printf("Wrong operator\n\n"); 
      } 
     } 
    } 

    return (0); 
} 

int childFunction(char *string) 
{ 
    int n1, n2; 
    char op; 
    float div = n1/n2; 

    printf("I am a child working for my parent\n\n"); 
    sscanf(string, "%d %c %d", &n1, &op, &n2); 

    if ((sscanf(&input,"%d %c %d", &n1, &op, &n2)) != 3) 
     exit(50); 

    if (op == '/' && n2 == 0) 
     exit(100); 

    switch (op) 
    { 
      case '+': 
      printf("\n%d %c %d = %d", n1, op, n2, n1+n2); 

      case '-': 
      printf("\n%d %c %d = %d", n1, op, n2, n1-n2); 

      case '/': 
      printf("\n%d %c %d = %f", n1, op, n2, div); 

      case '*': 
      printf("\n%d %c %d = %d", n1, op, n2, n1*n2); 

      default: 
      exit(200); 
    } 
    exit(0); 
    sleep(1); 
} 

Answer 1

char input; 

if ((sscanf(&input,"%d %c %d", &n1, &op, &n2)) != 3) 
    exit(50); 

的sscanf第一个参数需要一个指向字符串的指针，而不是单个字符。