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the word "stack" has been mentioned multiple times in this thread, would really appreciate if someone could explain in detail? does this have anything to do with compile-time allocation vs. run-time...

Salem, would you mind elaborate on that? what is this "stack" you were referring to?

this code snippet compiles and runs fine


#include <stdio.h>

int main(void) {
int i, a[i];


i=2;

Thank you!

so running the following code



#include <stdio.h>

int main(void) {
char c;
int i=1;

thank you! that worked, may i ask why it had to be an explicit comparison? wouldn't implicit conversion take care of the returning "int" value of scanf() to a TRUE, and only errors such as EOF to...

linux on a PC

Ctrl+Z just backgrounds it

so i'm using scanf() this way



int i;

while (scanf("%i", &i))
printf("%i ", i);
printf("\ndone\n");

Thank you all very much for your help!
anduril462, just want to say this, i'm telling myself as i type up this reply, that i shouldn't expect my future questions to be answered with this level of...

i apologize, the actual code i was compiling with has more lines to it, i thought i dump it down to simplify the question, could you try with these 2 code samples, the gcc version shows 4.4.7

...

so i had code "test.c" like this


#include <stdio.h>
#include <math.h>

int main(void) {
float f;
f=pow(2.1, 2);
printf("f is %f\n", f);

Thank you all for your help!

i kind of understand what you are saying :-) that "%f" is merely a formatting string, and that's how C does things, but still if you just read the above quoted, it's quite ambiguous, and that's the...

why doesn't integer to float conversion kick in here


printf("%f", 1)

it kicks in here


float f;
int x=3, y=2;

i'm trying to understand how the float/integer conversion mechanism works here, i intentionally made up the code like that, i was expecting "1.000000", in fact i had something like


float f;
f =...

why does the following code output "0.000000" instead of "1.000000"?


#include <stdio.h>

int main(void) {
int x=3, y=2;
printf("3/2 is %f\n", x/y);
return 0;
}