c - 我这样做是为了平衡方程的括号，但我认为这是错误的检查并纠正它

Question

int main(){
char i,input[30],close,open;
for(i=0;i<='.';i++){
  printf("enter equation");
  scanf("%c",input[i]);
  if(input[i]=='(')
    input++;
  input[i]=open;
  else if(input[i]==')')
    input[i]--;
  input[i]=close;
  else if(open[i]==close[i])
  {
    printf("parenthesis are balance");
  }
  else
    printf("parenthesis are not balance");
  }

  getch();
  return 0;
}

Answer 1

如果您在代码上使用缩进，您可以更容易地发现问题:

int main()
{
    char i,input[30],close,open;
    for(i=0;i<='.';i++) // why i <= '.'? maybe you mean int i and input[i] != '.'...
    {
        printf("enter equation");
        scanf("%c",input[i]); // you need &input[i]. In fact, I think what you need is scanf("%s", input); but outside of this for loop...
        if(input[i]=='(')
            input++; // Do you mean input[i]++?
        input[i]=open; // this isn't inside the if condition. Use brackets if you want it to be
        else if(input[i]==')') // won't compile because there's no matching if
            input[i]--;
        input[i]=close; // not inside the else. Also, what is close and open? You don't initialize them
        else if(open[i]==close[i]) // open and close are not arrays. You can't use them like this
        {
            printf("paranthesis are balance");
        }
        else
            printf("paranthesis are not balance");
    }

    getch();
    return 0;
}

有很多错误。我建议阅读教程。 This one ， 例如。您可以通过“C 教程”Google 搜索更多内容