CODEWITHSUNDEEP
ccastingdoubleprintf
user1121487
user1121487

Reputation: 2680

printf() and float error

int number1 = 23;
int number2 = 100;

printf("Output: %.2d", double(number1) / number2);

This I want:

Output: 0.23

This I get:

error: expected expression before ‘double’

I don't understand the error message. How to do cast the integer to double and perform the calculation?

Upvotes: 0

Views: 2007

Answers (5)

kotAPI
kotAPI

Reputation: 419

Let's have a look at some basics of format specifiers shall we?

**Format specifier**    **Characters matched**  **Argument type**

%c                           any single character   char

%d, %i                       integer            integer type

%u                           integer            unsigned

%o                           octal integer      unsigned

%x, %X                       hex integer        unsigned

%e, %E,%f, %g, %G            floating point 
                             number             floating type

%p                           address format     void *


%s                           any sequence of 
                             non-whitespace 
                             characters         char

You used "%d" instead of "%f" there is no way the compiler is gonna print a floating point value.

Upvotes: 1

Sadique
Sadique

Reputation: 22813

That's because you are typecasting the wrong way, also the format specifier is incorrect:

printf("Output: %.2f", (double)number1 / number2);

Here is the Output:

http://ideone.com/10a1Ka

Upvotes: 2

David Heffernan
David Heffernan

Reputation: 612794

You have two errors.

Firstly you are casting incorrectly. Cast to double like this:

(double)number1

Secondly you used the wrong print format specifier. You used d which is for integers. Use f instead. Putting it all together gives:

printf("Output: %.2f", (double)number1 / number2);

Upvotes: 1

Yu Hao
Yu Hao

Reputation: 122373

The correct syntax of cast is (double)number1 / number2, and you should use %.2f as the format specifier.

printf("Output: %.2f", (double)number1 / number2);
//                 ^   ^      ^

Upvotes: 1

Gangadhar
Gangadhar

Reputation: 10516

Use%f format specifier 

   printf("Output: %.2f", (double) number1 / number2);

Upvotes: 1

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