why %s works instead of %c in scanf("%c",y);

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when i used %c in scanf("%c",y)1 in my code. it didnt work. but when i replaced %c with %s. it worked. why? please answer asap. i have exam tomorrow and i need to defend it.
heres my entire code.

C++

#include<stdio.h>
#include<conio.h>

void main(){
int x; //sport
char y;//role,event,style,etc
clrscr();
printf("enter a number between 1-5:");
scanf("%d",&x);
if(x==1){
	printf("basketball\n");
	printf("enter a or b:");
	scanf("%s",&y);
	if(y=='A'||y=='a'){
	printf("forward");
	}
	else if(y=='B'||y=='b'){
	printf("guard");
	}
	else{
	printf("invalid input");
	}
}
else if(x==2){
	printf("badminton\n");
	printf("enter a or b:");
	scanf("%s",&y);
	if(y=='A'||y=='a'){
	printf("singles");
	}
	else if(y=='B'||y=='b'){
	printf("doubles");
	}
	else{
	printf("invalid input");
	}
}
else if(x==3){
	printf("baseball\n");
	printf("enter a or b:");
	scanf("%s",&y);
	if(y=='A'||y=='a'){
	printf("catcher");
	}
	else if(y=='B'||y=='b'){
	printf("pitcher");
	}
	else{
	printf("invalid input");
	}
}
else if(x==4){
	printf("swimming\n");
	printf("enter a or b:");
	scanf("%s",&y);
	if(y=='A'||y=='a'){
	printf("back stroke");
	}
	else if(y=='B'||y=='b'){
	printf("butterfly");
	}
	else{
	printf("invalid input");
	}
}
else if(x==5){
	printf("track and field\n");
	printf("enter a or b:");
	scanf("%s",&y);
	if(y=='A'||y=='a'){
	printf("discus throw");
	}
	else if(y=='B'||y=='b'){
	printf("100m dash");
	}
	else{
	printf("invalid input");
	}
}
else{
	printf("invalid input");
}
getch();
}

Posted 9-Sep-15 6:55am

Member 11971871

Updated 9-Sep-15 7:01am

PIEBALDconsult

v2

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Comments

PIEBALDconsult 9-Sep-15 13:10pm

"If you have to include code, include the smallest snippet of code you can." You would do well to write a very small test app that demonstrates the problem.
Also, I recommend that you write a function that handles input rather than repeating the code as you do -- it reduces the possibility of bugs and makes maintenance easier.

Member 11971871 9-Sep-15 13:28pm

thanks! i just started coding. i still dont know how to create a function. my only problem is the compiler doesnt read the scanf()... line inside the if(x==1), and after the skipping, it automatically execute the statement in else (inside the if(x==1)). the problem occurs while using %c.

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CPallini · Accepted Answer · 2015-09-09T07:19:00

Solution 1

You should never use

C

scanf("%s", &y);

if y is declared as char.

You should always check scanf function result (it should be 1 in your case).

You have to consider that console input is buffered, that is, scanf doesn't return until you press return.

Please consider posting more meaningful info, "It doesn't work" is really poor.

Posted 9-Sep-15 7:19am

CPallini

Comments

Sergey Alexandrovich Kryukov 9-Sep-15 13:23pm

Absolutely, a 5.
—SA

CPallini 9-Sep-15 15:34pm

Thank you.