CODEWITHSUNDEEP
cioprintf
Peter Olson
Peter Olson

Reputation: 142939

Why does getchar make a call to printf stop working?

I wrote a program that accepts a character of input and outputs that character, like this

int ch = getchar();
printf("%c", ch);

It worked like I expected. Then I decided to be welcoming and print Hello first.

printf("Hello!\n");
int ch = getchar();
printf("%c", ch);

To my surprise, this caused the compiler to throw two errors:

error C2065: 'ch' : undeclared identifier
error C2143: syntax error : missing ';' before 'type'

I didn't see why adding the first line would cause that to happen. Anyway, I refactored the program to get rid of the int declaration and the errors magically disappeared.

printf("Hello!\n");
printf("%c", getchar());

What's going on? What's the magic that causes these errors to appear and then disappear?

Upvotes: 3

Views: 1248

Answers (4)

Arthur Camara
Arthur Camara

Reputation: 577

Probably, you are using an pretty old C compiler. Probably one with C89 support. Using so forces you to declare any variable before anything within a block, (eg an function or main) Two ways out: Declare ch first:

int ch = getchar();
printf("Hello!\n");
printf("%c", ch);

or, even better, try changing your compiler. What OS are you using? Windows? Linux? Mac?

Also, a quick note. You are using getchar to get an integer number.

Try using, instead of getchar, scanf("%d", &ch). Or, if you really needs to use getchar, and to print it as a char, declare ch as a char itself, and, if you need to use it as an integer again, use the itoa function, that transforms a char to an integer.

Upvotes: 0

Alok Save
Alok Save

Reputation: 206546

Creating new variables after the start of a block was not allowed in C89 standard but is allowed in the newer C99 standard.

You are using a older compiler or a compiler not fully compliant to c99.
Your code example should work as is on any good compiler. Works on gcc-4.3.4

Alternate Solutions:

You can get rid of the problems in two ways:
Declare the variable at the begining of the block: 

int ch;
printf("Hello!\n");
ch = getchar();
printf("%c", ch);

Or

Create a new block for declaring the variable: 

printf("Hello!\n");
{ 
    int ch = getchar();
    printf("%c", ch);
}

Suggestion:

You should really change your compiler because if i remember correctly gcc supported this as compiler extension even prior to c99.

Upvotes: 3

dreamlax
dreamlax

Reputation: 95335

Versions of C prior to C99 did not allow “mixed declarations and code”, meaning you had to declare all of your variables at the beginning of the scope. Modern-day C compilers allow mixed declarations and code, as do C++ compilers. Some non-C99 compilers even allow it as an extension.

I assume this was to make it easier for a compiler to determine how much space would actually be required on the stack, or something along those lines.

Upvotes: 2

poy
poy

Reputation: 10507

If you're using an older c compiler , you have to make all variable declarations BEFORE anything else. Try:

int ch;
printf("Hello!\n");
ch = getchar();
printf("%c", ch);

Upvotes: 5

Related Questions