Code:main() { int i=3; printf("value of i=%d",i); printf("address of i=%u",&i); printf("value of i=%d",*(&i)); }
Code:main() { int i=3; printf("value of i=%d",i); printf("address of i=%u",&i); printf("value of i=%d",*(&i)); }
C enlightened
There are no error messages,I just cannot see the output screen.
C enlightened
stdout is buffered and will display when it reaches a newline or is flushed.
gg
Hello codeplug,can you explain what is stdout?Is it a pre processor directive like stdio?
C enlightened
Thanks,got the output,just had to use a pre-processor directive.
C enlightened
>> can you explain what is stdout?
"stdout" is the stream that printf() prints to.
gg
The only preprocessor directive needed is #include <stdio.h>Originally Posted by sameertelkar
If you do this you will get an output:
Code:#include <stdio.h> main() { int i=3; printf("value of i=%d",i); printf("address of i=%u",&i); printf("value of i=%d\n",*(&i)); return; }
As long as we are making improvements to the code, we may as well fix everything:
Code:#include <stdio.h> #include <stdlib.h> int main(void) { int i = 3; printf("value of i = %d\n", i); printf("address of i = %p\n", (void*)&i); printf("value at %p = %d\n", (void*)&i, *(&i)); return EXIT_SUCCESS; }
