printf in gcc wrong result

Question

Case 1: printf("%f",(7/2)); in gcc output is 0.000000.

Case 2: float k= 7/2; printf("%f",k); in gcc output is 3.000000.

In the first case printf expects float but gets Integer so gives wrong result. But in second case it does type conversion.

Here are my questions-

1. Why does not gcc give type mismatch error/ warning in the first case?
2. In 2nd case it is doing type conversion by default but why not in 1st case?

Answer 1

In 2nd case it is doing type conversion by default but why not in 1st case?

In first case 7 and 2 both are of int type. Dividing 7 by 2 will give you an int . Printing it with %f will invoke undefined behavior . You will get anything. In this case there is no type conversion.

Try this

printf("f", (7.0/2));  

In second case k is of float type hence the result of 7/2 is converted to the type of k by default.

Why does not gcc give type mismatch error/ warning in the first case?

Compiling the first statement with -Wall flag is giving the warning:

[Warning] format '%f' expects argument of type 'double', but argument 2 has type 'int' [-Wformat=]

Answer 2

1) gcc does not by default check types in format string with arguments - so it "interpretes" the data (ie. the integer) as float at runtime

2) gcc now interpretes the float as float at runtime .

Answer 3

GCC can give a warning, but you didn't enable it. Try compile with flag -Wall , you would see:

warning: double format, different type arg (arg 2)