#include<iostream> #include<stdio.h.> using namespace std; int main() { float f=11.11; printf("%d",f); } When i execute the following code in dev c++ it gives the output -536870912. When i execute the same code on online tutorials point compiler i get different on every run. What could be the reason behind?
If the argument doesn't match the format specifier, it's undefined behaviour. Ue %f to print a float.
printf("%f",f); You should be able to catch these sort of errors easily with good compilers. GCC produces:
warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘double’ [-Wformat=] for your code.
P.S.: There's a stray dot in your stdio.h header inclusion.
I'm not sure whether you're asking Why did it not print 11? or Why did it print different answers under different compilers?.
The reason it didn't print 11 is that, as other answers have explained, %d doesn't print floats, and in printf calls there are no automatic conversions to fix up mismatches between passed and expected types.
And the reason it printed different things under different compilers is that it's undefined behavior, meaning anything can happen. Suppose you go out to your car, loosen all the lugnuts on the front wheel until it's about to fall off, get in, and drive down the road at high speed until the wheel does fall off, whereupon you lose control and crash into a ditch. And suppose you get a new car and do exactly the same thing tomorrow, except that by chance you crash into a tree instead. At this point there are two approaches you could take:
%c is defined as accepting an int. Also because there are some automatic conversions that do happen to the arguments when you call printf: char and short are promoted to int, and float is promoted to double. These are called the usual argument promotions or something like that, and they apply for any function that (like printf) accepts a variable number of arguments, and so has no fixed prototype.It's a float, use
printf("%f" , f); If you don't it's an undefined behaviour. To understand you can try to cast a int with value -1 to an unsigned int.
You will get an huge value.
f in fact expects a double not a float since C99 at least.
double is promoted to double for varargs arguments it doesn't really matter.There is a mismatch between the conversion specifier and the type of the variable here:
printf("%d",f); Passing to printf other parameter(s) then specified by the conversion specifier(s) given by the format string provokes undefined behaviour. From that moment on everything can happen.
To fix this use the correct conversion specifier(s).
For floating point variables use the conversion specifier f, which expects a double. However when a float is passed to a variadic function like printf the float gets promote to a double.