Thread: Scanf for char doesn't work in loop

What am I doing wrong here? I've tried this in the main, I've tried it in a function using a return statement, and here is my attempt to pass a pointer to a character (a math operator--this is supposed to be a calculator program) into a loop. It goes in the first time, but then after that it ignores the input of the character and just prints out the statements without waiting for input. Can you look at this program fragment and tell me what to do to fix it?

Thanks a lot.
Donna

#include <stdio.h>

void get_operator(char *lop);
int get_operand();

void main(){
// int num1=0;
char op;
int num2=0;
// int accum=0;
int i;

for (i =1; i < 4; ++i){
get_operator(&op);
printf("the operator is %c\n", op);

num2 = get_operand();
printf("operand is %d\n", num2);
}
printf("the final result is %d\n", num2);
}

void get_operator( char *lop){
printf("enter a math operator or q to quit> ");
scanf("%c", lop);
}
int get_operand(){
int lnum2; /* local variable for operand */
printf("enter an integer >");
scanf("%d", &lnum2);
return lnum2;
}
}

AAAAAAAAHHHH
scanf(), I never use it, it's dangerous(has a tendency to leave newline characters in the buffer) use getch() instead or for numeric data use gets() and then sscanf()

The & in front of the variable made no difference. It still ignored the thing the 2nd time through.

Tried using getc--but it said that this was only for unions or structures. I'm just trying to enter one little character!

What is the syntax for getc when reading from direct input, not from a file?


Donna

try
ch = getc(stdin);
or
ch = getchar();
or
ch = getch();
the last one needs <conio.h>

Thanks for all your suggestions.

Just wanted to let you know that my instructor told me the solution is to put a space in front of the character reference in the scan statement:
scanf(" c", &lop); (or in this case, just lop since it is a pointer already)

Huh!

Donna

Put this after the scan:

while(getchar() != '\n') continue;

clears the standard input buffer.

Originally posted by SPOOK
try
ch = getc(stdin);
or
ch = getchar();
or
ch = getch();
the last one needs <conio.h>

getch() is not ansi C , one cant expect everyone to have a conio.h in their implementation

Originally posted by Engineer
Putting space after before the % does the trick. I once ran into this in one of my first C assignments.

scanf(" %c", &mychar);

The reason this works is that the first space flushes stdin and then lets the user input the character.

Cheers.

It may work but then again it should not
from K&R
"The format string may contain
.Blanks and and tabs which are ignored
.ordinary characters(not %)which are expect to match the next non white-space character of the input stream.
.conversion specifications , consisting of a % ,an optional assignment supression character * ,an optional number specifying the field width ,an optional h,l or L indicating the width of the target , and a conversion character
"

So for a ansi compliant compiler scanf("%c",&mychar); and
scanf(" %c",&mychar) should lead to the same effect as blanks and tabs are supposed to be ignored .

In my best opinion, avoid scanf. I was recently having problems with it and I came to the conclusion that there are numberous ways to get around it. For instance, the ol' fgets and sscanf routine. That works great.

--Garfield