printf statement not executed

[JS]

Why is "TEST" not printed when I run this code after comiling?


#include <stdio.h>

int main()
{

int a,b,c,d,e,x;
int *variables[5];

a = 1;
b = a*2;
c = b*2;
d = c*2;
e = d*2;

variables[0] = &a;
variables[1] = &b;
variables[2] = &c;
variables[3] = &d;
variables[4] = &e;


for(x=0; x<5; x++)
printf("Variable %c = %d\n", 'a'+ x, *variables[x]);


printf("TEST");


return(0);
}

 

[Ken Human]

Why is "TEST" not printed when I run this code after comiling?


#include <stdio.h>

int main()
{

int a,b,c,d,e,x;
int *variables[5];

a = 1;
b = a*2;
c = b*2;
d = c*2;
e = d*2;

variables[0] = &a;
variables[1] = &b;
variables[2] = &c;
variables[3] = &d;
variables[4] = &e;


for(x=0; x<5; x++)
printf("Variable %c = %d\n", 'a'+ x, *variables[x]);


printf("TEST");


return(0);
}

I ran this code unmodified, "TEST" printed on my console.

 

[Walter Roberson]

The OP missed the \n after TEST, so the output was probably
wiped out by the command prompt.

 

[Jonathan Bartlett]

Why is "TEST" not printed when I run this code after comiling?

Because you did not add a "\n" or flush your output. The "f" family of
functions does buffering, and may not print out all of your results
until it hits a "\n" or you tell it to flush.

The printf is executed, it's just that your "TEST" is waiting in a
buffer and hasn't yet been handed to the operating system.

Jon

 

[Randy Howard]

Why is "TEST" not printed when I run this code after comiling?
printf("TEST");

Try putting a \n in there.

 

[Mark McIntyre]

int *variables[5];

as an aside, you do know that this creates an array of pointers to
ints?

variables[0] = &a;

to be fair, you use it as such. However I'm wondering whether you
don't realise why.

 

[Martin Ambuhl]

Why is "TEST" not printed when I run this code after comiling?

If you don't terminate the last line of output with an end-of-line
character, there is no portable behavior.

[...]

 

[Lawrence Kirby]

Because you did not add a "\n" or flush your output. The "f" family of
functions does buffering, and may not print out all of your results
until it hits a "\n" or you tell it to flush.

The program does flush the output by terminating normally. On normal
program termination all open files are flushed (when that is a meaningful
operation) and closed.

The printf is executed, it's just that your "TEST" is waiting in a
buffer and hasn't yet been handed to the operating system.

If an implementation did that it would be broken. It is true that if
the last line is not terminated by a newline it may not be
considered valid and perhaps even discarded but this is not a
buffering issue. In practice I'm not aware of an implementation that would
discard it. Walter probably has the correct explanation.

Lawrence

 

[Villy Kruse]

The program does flush the output by terminating normally. On normal
program termination all open files are flushed (when that is a meaningful
operation) and closed.

It has been seen before that the output that was actualy written was
overwritten by the next command prompt, and people has in that case
spent hours looking for a bug that wasn't there. Including a newline
would have forced the prompt to the next line, though.


Villy

 

[Lawrence Kirby]

It has been seen before that the output that was actualy written was
overwritten by the next command prompt, and people has in that case
spent hours looking for a bug that wasn't there. Including a newline
would have forced the prompt to the next line, though.

Yes, there was a reference to this in the part of my article that you
snipped, i.e. Walter's explanation.

Lawrence