C Simple calculator, getting rid of useless lines - c

Question

The calculator program was an assignment for my C introduction web class.
How would I better include this section: (currently included in every else if statement)
printf("\nSyötä ensimmäinen luku:");
scanf("%d", &ekaluku);
printf("Syötä toinen luku:");
scanf("%d", &tokaluku);
in the code without changing output?
Source code:
#include <stdio.h>
int laskesumma(int a, int b);
int laskeerotus(int a, int b);
int lasketulo(int a, int b);
int main(void)
{
int valinta, ekaluku, tokaluku;
do {
printf("1: kahden luvun summa\n2: kahden luvun erotus\n3: kahden luvun tulo\n<0: ohjelman lopetus\n\nValitse laskutoimitus:");
scanf("%d", &valinta);
if (valinta < 0)
{ printf("Ohjelma lopetetaan...\n\n"); }
else if (valinta == 1)
{
printf("\nSyötä ensimmäinen luku:");
scanf("%d", &ekaluku);
printf("Syötä toinen luku:");
scanf("%d", &tokaluku);
printf("%d + %d = %d\n\n", ekaluku, tokaluku, laskesumma(ekaluku,tokaluku));
}
else if (valinta == 2)
{
printf("\nSyötä ensimmäinen luku:");
scanf("%d", &ekaluku);
printf("Syötä toinen luku:");
scanf("%d", &tokaluku);
printf("%d - %d = %d\n\n", ekaluku, tokaluku, laskeerotus(ekaluku,tokaluku));
}
else if (valinta == 3)
{
printf("\nSyötä ensimmäinen luku:");
scanf("%d", &ekaluku);
printf("Syötä toinen luku:");
scanf("%d", &tokaluku);
printf("%d * %d = %d\n\n", ekaluku, tokaluku, lasketulo(ekaluku,tokaluku));
}
else
{ printf("Antamasi luku ei kelpaa.\n"); }
} while (valinta>0);
return 0;
}
//Addition
int laskesumma(int a, int b) {
int tulos;
tulos = a+b;
return tulos;
}
//Subtraction
int laskeerotus(int a, int b) {
int tulos;
tulos = a-b;
return tulos;
}
//Multiplication
int lasketulo(int a, int b) {
int tulos;
tulos = a*b;
return tulos;
}
Is there anything else I could fix for better readability or decreasing the size of the program?

Answer 1

One possibility is:
int ekaluku_tokaluku(int *ekaluku, int *tokaluku)
{
printf("\nSyötä ensimmäinen luku:");
if (scanf("%d", ekaluku) != 1)
return EOF;
printf("Syötä toinen luku:");
if (scanf("%d", tokaluku) != 1)
return EOF;
return 0;
}
Note that the variables are passed to the function as pointers, so there is no & in the scanf() calls. Some people might use a prefix such as p_ekaluku (or pEkaluku) on the names to indicate that they're a pointer.
You'd use it like this:
else if (valinta == 1)
{
if (ekaluku_tokaluku(&ekaluku, &tokaluku) != 0)
break;
printf("%d + %d = %d\n\n", ekaluku, tokaluku, laskesumma(ekaluku, tokaluku));
}
Returning 0 on success is a common pattern. It's always worth checking that your code handles EOF conditions. You could consider adding fflush(stdout); after the printf() and before the scanf() statements in the function.

Answer 2

Just use a function if you're repeating those lines of codes a lot and return the integer value.
int userInput() {
int ekaluku;
printf("\nSyötä ensimmäinen luku:");
scanf("%d", &ekaluku);
return ekaluki;
}

Answer 3

Whenever you repeat a code block somewhere in your program, you should always consider to put that code block into a function. That will make it much easier to maintain your code.
Besides that I'll suggest that you don't use scanf - it has many surprises. For instance try to give your program the input 1a
Instead use sscanf and fgets as they'll give you less surprises.
The function could look like:
int get_input(FILE *fp, int n)
{
int d = 0;
char line[LINESIZE];
if (n > 0)
{
printf("Input the %d. number\n", n);
}
if(!fgets(line, LINESIZE,fp))
{
printf("Error reading input\n");
return 0;
}
if (sscanf(line, "%d", &d) != 1)
{
printf("Error reading input\n");
return 0;
}
return d;
}
and it can be used like:
do
{
printf("1: kahden luvun summa\n2: kahden luvun erotus\n3: kahden luvun tulo\n<0: ohjelman lopetus\n\nValitse laskutoimitus:");
valinta = get_input(stdin, 0);
if (valinta < 0)
{
printf("Ohjelma lopetetaan...\n\n");
}
else if (valinta == 1)
{
ekaluku= get_input(stdin, 1);
tokaluku= get_input(stdin, 2);
printf("%d + %d = %d\n\n", ekaluku, tokaluku, laskesumma(ekaluku,tokaluku));
}

Answer 4

This can be a little bit shorter (number of lines):
int laskesumma(int a, int b) { return (a+b);} //Addition
int laskeerotus(int a, int b){ return (a-b);} //Subtraction
int lasketulo(int a, int b) { return (a*b);} //Multiplication
but lower readibility in case of needed more complicated functions...

Answer 5

reduce code like this
#include <stdio.h>
#define make_func(name, op)\
static inline int name(int a, int b){\
return a op b;\
}\
/**/
make_func(laskesumma , +) //Addition
make_func(laskeerotus, -) //Subtraction
make_func(lasketulo , *) //Multiplication
#define op(no) ( " +-*"[no] )
typedef int (*Func)(int,int);
#define func(no) ( (Func[]){ 0, laskesumma, laskeerotus, lasketulo}[no] )
enum { QUIT, ADD, SUB, MUL };
int main(void){
int valinta, ekaluku, tokaluku;
do {
printf("%d: kahden luvun summa\n"
"%d: kahden luvun erotus\n"
"%d: kahden luvun tulo\n"
"<%d: ohjelman lopetus\n\n"
"Valitse laskutoimitus:", ADD, SUB, MUL, QUIT);fflush(stdout);
scanf("%d", &valinta);
switch(valinta){
case ADD:
case SUB:
case MUL:
printf("\nSyota ensimmainen luku:");fflush(stdout);
scanf("%d", &ekaluku);
printf("Syota toinen luku:");fflush(stdout);
scanf("%d", &tokaluku);
printf("%d %c %d = %d\n\n", ekaluku, op(valinta), tokaluku, func(valinta)(ekaluku, tokaluku));
break;
default:
if(valinta < QUIT){
printf("Ohjelma lopetetaan...\n\n");
return 0;
}
printf("Antamasi luku ei kelpaa.\n");
}
} while (valinta > QUIT);
return 0;
}

Answer 6

Try to use switch case in your code
do
{
if(valinta<0)
printf("Ohjelma lopetetaan...\n\n");
switch(valinta)
{
case 1:
case 2:
case 3:
printf("\nSyötä ensimmäinen luku:");
scanf("%d", &ekaluku);
printf("Syötä toinen luku:");
scanf("%d", &tokaluku);
break;
default: printf("Antamasi luku ei kelpaa.\n");
}
}while(valinta>0);