How to use while with two situations?

Question

I use while(ch!=3 || ch!=17) to run the program, but when I press "ctrl+q" or "ctrl+c", it dosen't stop. How to correct it? Thank you.
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#include <stdio.h>
#include <stdlib.h>
int main(void)
{
char ch;
while(ch!=3 || ch!=17)
{
ch=getche();
printf("\nASCII of ch=%d\n",ch);
}
system("pause");
return 0;
}

Answer 1

Your original while should be:
while (ch!=3 && ch!=17)

If ch is 48 or '0', it is not equal to both. When it is cntrl-c, it is 3. But, it is not 17 so the loop keeps going. The only way to stop that loop was for ch to be both 3 and 17.

Answer 2

See John Wells knew more about the codes than I did. Plus I missed that fatal piece of logic where an OR in a while loop for key pressing is not ok. It was late (excuses), but I'm glad you got some actual help

Answer 3

John Wells Zeke Williams Thank you very much !!

Answer 4

I don't know the getche function that well, but I thought it only waits for ONE single character. 17 is two. Are you trying to check for the character code? If so, you'll have to convert ch to an integer. Otherwise, you need to put single quotes around 3. 17 won't work as a character: != '3'

Answer 5

Zeke Williams
Thank you for your help! But I thought it is not the problem about numbers of character. When I used while(ch!=17) or while(ch!=3) individually, it works. I want to know why put them together with "||" doesn't work in while function.
Btw, I modified the program(below) successfully and it works.

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#include <stdio.h>
#include <stdlib.h>
int main(void)
{
char ch;
int a;
while(a!=1)
{
ch=getche();
printf("\nASCII of ch=%d\n",ch);
if (ch==3 || ch==17)
a=1;
}
system("pause");
return 0;
}