Why printf(“%.1f”, 1) output 0.0

Question

I use the codeblock .

when the code is:

printf("%.1f", 1);

The program can run,and the output is 0.0 . I want to know why. `

Answer 1

Change it to:

printf("%.1f", 1.0);

f conversion specifier requires an argument of type double but you are passing an int value ( 1 is of type int ). Passing an argument of the wrong type to printf invokes undefined behavior.

Answer 2

Using wrong format specifier invokes undefined behavior . You may get either expected or unexpected result. Use %d instead as the argument passed to printf is int type or change 1 to 1.0 if you are using %f .

C11: 7.21.6 (p9):

If a conversion specification is invalid, the behavior is undefined. ²⁸²⁾ If any argument is not the correct type for the corresponding conversion specification, the behavior is undefined.

Answer 3

The format is incorrect , you should give a float or a double as the second argument to the printf function. The gcc compiler also gives a warning on such mistakes by programmers.

 printf("%.1f",1.23);

Output:

 1.2

Be carefull with these kind of mistakes. Good luck!