:::::::::::::::::::::::::::: c program :::::::::::::::::::::::::: #include void main() { char ch=90; printf("%c",(char*)ch); } what would be the output & why ?

Question

:::::::::::::::::::::::::::: c program ::::::::::::::::::::::::::
#include<stdio.h>
void main()
{
char ch=90;
printf("%c",(char*)ch);
}
what would be the output & why ?

anonymous · Answer

output is "Z".

90 is Ascii value of "Z"(caps Z).

anonymous · Answer

Hope you got the answer.

anonymous · Answer

what is your expected output?

anonymous · Answer

i want to know why this is happening ???

anonymous · Answer

because you are defining %C .
printf("%c",(char*)ch); OUTPUT WILL BE "Z"
printf("%d",(char*)ch); OUTPUT will be "99"

anonymous · Answer

when char ch=90 executes then what gets stored in ch ?

anonymous · Answer

90

anonymous · Answer

and when printf("%c",(char*)ch); executes then what happens ?

anonymous · Answer

It prints it as character. because %c is indication of char value.

anonymous · Answer

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