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#define SEM 59 //59 is ASCII value of semicolon

void main()
{
if(printf("%c",SEM))
{
}
}

if you know the unicode of ;, you can print..

but unicode of ; was nt given

can be printed using ASCII value for semicolon.

If u don't remember the ASCII value of semicolon then write a program to find it.Its just 3 line of code:

int ch;
for(ch=1;ch<=256;ch++)
printf("%d \t %c \n",ch,ch);

which I think would not be tough.

@kk it was written test.. how can i find value of semi colon on paper.

int main()
{
if(int var=59)
if(printf("%c",var))
{
}
}

I think we have to access the text segment in the runtime and print the ";" in that segment.

i guess you can use the ascii values to print ';' which i my case is 59, here is the code to find it.

int i;
for(i=0; i< 255; i++)
printf(" %c-%d", i, i);

Using the ASCII value of the semicolon, print it as a char.

main()
{
if(printf(";"))
}

#include<stdio.h>
main()
{
if(printf(;))
{}
}

#include<stdio.h>
main()
{
if(printf(";"))
{}
}