scanf（“％C”，& opt）和scanf（“％C”，& opt）之间的差异

问题描述

我在IDE TurboC3中执行我的c语言程序。

非工作代码如下

I am executing my c language program in IDE TurboC3.
Non working code is as follows

#include<stdio.h>
#include<conio.h>
void main()
    {
    int i, j, t;
    char opt;
    clrscr();
    printf("Enter two values" );
    scanf("%d", &i);
    scanf("%d", &j);
    printf("\n\n press + for sum\n");
    printf("press * for mult\n");
    printf("print - for subtract\n\n");
    scanf("%c", &opt); // code that is creating problem //
    switch(opt)
        {
        case '+':
            {t=i+j;
            printf("\nsum =%d", t);
            }
            break;
        case '*':
            {t=i*j;
            printf("\nproduct =%d", t);
            }
            break;
        case '-':
            {t=i-j;
            printf("\ndifference =%d", t);
            }
        break;
        default:
            printf("\ninvalid choice");
        }
    getch();
    }

代码块已更正，缩进添加 - OriginalGriff [/ edit]



我尝试了什么：



帮助我执行代码的更正如下

[edit]Code block corrected, indentation added - OriginalGriff[/edit]

What I have tried:

The correction that helped me execute the code is as follows

scanf(" %c", &opt); // code without problem //

即我在和％c之间插入了一个空格和代码工作。



请帮我理解其中的区别。

i.e. i have inserted a space between " and %c and the code worked.

Please help me understand the difference.

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