Question

我试图在printf声明显示中制作此if，但事实并非如此。以下是代码，我突出显示了printf部分。

char conv;
float cel;

printf("Enter what you'd like converted: \n\t1.Celcius to Fahrenheit\n\t2.Inch to CM\n\t3.CM to Inch\n");
scanf_s("%c", &conv);

// This one
if (conv == '1')
    printf("Please enter amount of Celcius: \n");
scanf_s("%f", &cel);

float fahr;
fahr = 33.8 * cel;

printf("Conversion of %f Celsius is %f Fahrenheit", cel, fahr);

getch();
return 0;

这是什么原因？

Answer 1

In looking at the line: scanf_s("%c", &conv);  
I see the function scanf_s(), 
which in not in any C library that I'm familiar with.

However, the function scanf(), which is in the C libraries, 
has the following prototype and usage:

int scanf(const char *format, ...)

where the return value is the number of items in the variable parameter list '...'
that were actually filled from the input.

I.E. your code should, after the call to scanf_s() check the return code
those value should be 1 in your example.

Then you have this code:

if (conv == '1')
    printf("Please enter amount of Celcius: \n");
scanf_s("%f", &cel);

supposedly with the idea that all those following lines would be 
enclosed in the 'if' block.
however, only the first line following the 'if' is actually part of the
'if' block.
To correct this problem, write the code using braces to delineate
the 'if' block, like this:

if (conv == '1')
{
    printf("Please enter amount of Celcius: \n");
    scanf_s("%f", &cel);

    float fahr;
    fahr = 33.8 * cel; // BTW: this is not the correct conversion formula

    printf("Conversion of %f Celsius is %f Fahrenheit", cel, fahr);
}

C - printf inside＆＃34; if＆＃34;声明没有显示

1 个答案: