Is the code below converting a character into its ASCII value?

Is the code below converting a character into its ASCII value?.

I faced a piece of code while studying evaluation of postfix operation,where it says "the expression converts a single digit character in C to its numerical value".?

in开发者_开发技巧t x=getch();  
int c=x-'0';      /*does c have the ASCII value of x?*/
printf("%d",c);  

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No, it's converting the ASCII value to a number >= 0.

Let's say you type '1'. getch() will return 49 which is the ASCII value of '1'. 49 - '0' is the same as 49 - 48 (48 being the ASCII value for '0'). Which gives you 1.

Note that this code is broken if you enter a character that is not a number.
E.g. if you type 'r' it will print 'r' - '0' = 114 - 48 = 66

(Ref.)

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No, it's giving the numeric value of a digit. So '5' (53 in ASCII) becomes 5.

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Is the code below converting a character into its ASCII value?

It isn't. It's doing the opposite (converting an ASCII value to the numerical value) and it only works for decimal digits.

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To print the ascii value all you need to do is :

int x=getch();  
printf("%d",x); 

If you are sure that you only want to accept integers as input then you need to put some constraints to the input before proceeding to process them.

int x = getch();
if (x >='0' || x <= '9') {
    int c = x - '0'; // c will have the value of the digit in the range 0-9 
    . . .     
}

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Any character(in your case numbers) enclosed within single quotes is compiled to its ASCII value. The following line in the snippet above translates to,

int c=x-'0'; ---> int c= x-48; //48 is the ASCII value of '0'

When the user inputs any character to your program, it gets translated to integer as follow,

If x = '1', ASCII of '1' = 49, so c= 49-48 = 1

If x = '9', ASCII of '9' = 57, so c= 57-48 = 9 and so on.