multiple getchar() in a row

[MareoRaft]

Here is a simple program which gives me unexpected results:

int main()
{
		
printf("\nTo calculate a^b, enter value a now.\n");
int a = getchar() - '0';
printf("\nEnter b now.\n");
int b = getchar() - '0';
printf("a = %d, b = %d\n\n", a, b);
		
return 0;	
}

if i hit a digit for a, then hit enter, it assigns -38 as the value of b.

the only way it works is if i enter a and b right next to each other, such as 52. this is not how i want it to behave. your advice greatly appreciated!

[Narue]

When you run the program, watch your hands carefully and count how many keys you hit. I suspect you're assuming that the Enter key is somehow treated differently when in reality it too is extracted by getchar as the '\n' character. -38 is fully expected when '\n' is 10 and '0' is 48 (as it is in ASCII).

[MareoRaft]

When you run the program, watch your hands carefully and count how many keys you hit. I suspect you're assuming that the Enter key is somehow treated differently when in reality it too is extracted by getchar as the '\n' character. -38 is fully expected when '\n' is 10 and '0' is 48 (as it is in ASCII).

thank you! i guess my problem now is that getchar() doesn't behave the way i'd like it to. Perhaps there is a different function I can use so that the user enters a digit for a, then the text "Enter b now." comes up, then they can enter digit b? I have tried getche() because it requires no ENTER, but now I receive the error message:

Undefined symbols:
"_getche", referenced from:
_main in cchtsT6q.o
_main in cchtsT6q.o
ld: symbol(s) not found
collect2: ld returned 1 exit status

thanks again! ;)

P.S. my code includes 1 line above "int main(){", and that is "#include <stdio.h>"
P.P.S i read somewhere that stdio.h defines getche() but does not actually have the function written out. If this is the case, is there some generic code or a new library I can include in my program to fix this?

[Narue]

Dude, just call getchar three times. It's not rocket science:

printf("\nTo calculate a^b, enter value a now.\n");
int a = getchar() - '0';
getchar(); /* Throw away the newline */
printf("\nEnter b now.\n");
int b = getchar() - '0';
printf("a = %d, b = %d\n\n", a, b);

To be truly robust you need to do more, but I fear I'll confuse you too much by explaining all of the nuances of stream I/O.

[MareoRaft]

I'm actually quite interested in a robust method if you don't mind. I want to build good habits as a programmer.

Also, I recognize that I am retrieving a character when really i should be using a function to retrieve a number, so that i won't be limited to the 10 digits.

I want people running this program in a Windows C compiler to be able to see the output before their window closes, so I added the line "std::cin.get();" before "return 0;", but now I receive the error message:

^[[A1.8.c: In function ‘main’:
1.8.c:24: error: expected expression before ‘:’ token

thanks,
Mareo

P.S. so my code now looks like this:

#include <stdio.h>
	
int main(int argc, char *argv[])
{

printf("\nTo calculate a^b, enter value a now.\n");
int a = getche() - '0';
printf("\nEnter b now.\n");
int b = getche() - '0';
printf("a = %d, b = %d\n\n", a, b);

std::cin.get();
return 0;	
}

[WaltP]

I'm actually quite interested in a robust method if you don't mind. I want to build good habits as a programmer.

Good habits as a programmer start with understanding the functions you are using and using them properly to get the results you want -- not by searching for a function that does exactly what you wish.

The most robust method is processing each key individually and doing exactly what you need to do with each character entered.

I want people running this program in a Windows C compiler to be able to see the output before their window closes, so I added the line "std::cin.get();" before "return 0;", but now I receive the error message:

Isn't that C++? Why would you add a C++ function to a C program? And one that is identical to getchar() too?

[[Alpha]-0mega-]

Since giving the solution right away seems to be frowned upon, I guess I'll give a few hints.

There are multiple ways to make this behave a bit more the way you want it to:

One uses a different approach for which I will give the following hints:
#include <stdlib.h>
use char array[] (which is a string)

Then there's two ways:
1.) replace getchar() with fgets()
2.) Use a while loop for getchar()
Use the magic command called ????

Profit!

The behaviour still isn't optimal but it should be a lot more like the way you want it to be.

Of course the main thing lacking in your program is the actual x^y calculation. I take it you know how to do that?

[MareoRaft]

just tried using fgets() (see below) but I run into message "warning: passing argument 1 of ‘fgets’ makes pointer from integer without a cast"

char a;
fgets (a, 1, stdin);

then i discovered fgetc() (see below), which works. unfortunately, works exactly the same as getchar().

int a = fgetc(stdin) - '0';

As Narue suggested, I am throwing out the value (see below).

printf("\nTo calculate a^b, enter value a now.\n");
int a = getchar() - '0';
printf("\nEnter b now.\n");
int b = getchar() - '0';
if( b == -38 ) b = getchar() - '0';
printf("a = %d, b = %d\n\n", a, b);

This is a method of using a while loop (see below), although I'm not sure if this is what you were hinting at.

printf("\nTo calculate a^b, enter value a now.\n");
int a = getchar() - '0';
printf("\nEnter b now.\n");
while ( getchar() != 10 );
int b = getchar() - '0';
printf("a = %d, b = %d\n\n", a, b);

not sure what you meant about a magic command. i couldn't find mention of any "profit()". perhaps i misunderstood you :P so as far as efficiency goes, getchar() is the way to go?

[Luther von Wulf]

just tried using fgets() (see below) but I run into message "warning: passing argument 1 of ‘fgets’ makes pointer from integer without a cast"

fgets() requires an array of char for the first argument.

#include <stdio.h>

int main() {
    char input[BUFSIZ];
    printf("\nTo calculate a^b, enter value a now.\n");
    int a = *fgets(input, BUFSIZ, stdin) - '0';
    printf("\nEnter b now.\n");
    int b = *fgets(input, BUFSIZ, stdin) - '0';
    printf("a = %d, b = %d\n\n", a, b);
    return 0;
}

unfortunately, works exactly the same as getchar().

Yes, getchar() is implemented in terms of fgetc(). The best way using getchar() is a loop until the '\n' character is found. The '\n' character in an interactive stream means that the user typed the return key.

#include <stdio.h>

void iflush(FILE* istrm) {
    int c;
    while ((c = getc(istrm)) != '\n' && c != EOF)
        ;
}

int main() {
    printf("\nTo calculate a^b, enter value a now.\n");
    int a = getchar() - '0';
    iflush(stdin);
    printf("\nEnter b now.\n");
    int b = getchar() - '0';
    printf("a = %d, b = %d\n\n", a, b);
    return 0;
}

[[Alpha]-0mega-]

just tried using fgets() (see below) but I run into message "warning: passing argument 1 of ‘fgets’ makes pointer from integer without a cast"

char a;
fgets (a, 1, stdin);

then i discovered fgetc() (see below), which works. unfortunately, works exactly the same as getchar().

int a = fgetc(stdin) - '0';

As Narue suggested, I am throwing out the value (see below).

printf("\nTo calculate a^b, enter value a now.\n");
int a = getchar() - '0';
printf("\nEnter b now.\n");
int b = getchar() - '0';
if( b == -38 ) b = getchar() - '0';
printf("a = %d, b = %d\n\n", a, b);

This is a method of using a while loop (see below), although I'm not sure if this is what you were hinting at.

printf("\nTo calculate a^b, enter value a now.\n");
int a = getchar() - '0';
printf("\nEnter b now.\n");
while ( getchar() != 10 );
int b = getchar() - '0';
printf("a = %d, b = %d\n\n", a, b);

not sure what you meant about a magic command. i couldn't find mention of any "profit()". perhaps i misunderstood you :P so as far as efficiency goes, getchar() is the way to go?

The hints I gave suggested you use a character *array*, that is:
char a[10]; (Where 10 can be any number)
However to use your 'math program' to produce the result of a^b, you will need to convert that char array to an integer when you're done. (And you still need to add the magic a^b formula)

Hence why I suggested using #include <stdlib.h> so *really* look up the command reference for the stdlib.h library

#include <stdio.h>
#include <stdlib.h>

#define MAX_DIGITS 9
/* If you want to use getchar() add i as well for the while loop */
int a,b; 
char aChar[MAX_DIGITS];
char bChar[MAX_DIGITS];

int main(void) {
printf("\nTo calculate a^b, enter value a now.\n");
fgets(aChar,MAX_DIGITS,stdin);
/* Do something here so that bChar does not get influenced by you pressing enter */
printf("\nEnter b now.\n");
fgets(bChar,MAX_DIGITS,stdin);
a = /*Convert aChar char array to int*/
b = /*Convert bChar char array to int*/
printf("a = %d, b = %d\n\n", a, b);
/* Actually calculate a^b here instead of just printing a = a and b = b (lookup math.h or use a for loop) */
getchar();
return 0;

[MareoRaft]

thanks a lot guys. i learned a lot from this :P