CODEWITHSUNDEEP
ctype-conversion
pratik jain
pratik jain

Reputation: 49

will type conversion take place in given code?

void main()
{
printf("%f",12/7.0);
getch();
}

Will the type conversion take place in this code and if yes then why it is taking place and if no why it is not taking place in this code please explain? Also I think 7.0 which is of double data type here (according to rule of type conversion) and 12 which is of integer data type would give data type double but when I print with %lf then the output on screen is again float I don't know how please correct me ?

Upvotes: 0

Views: 62

Answers (1)

Jean-Baptiste Yunès
Jean-Baptiste Yunès

Reputation: 36391

You can't deduce the type of the expression by the result of the printing. printf may show you only part of the value...

man printf says for f or F format (emphasize is mine):

fF

The double argument is rounded and converted to decimal notation in the style [-]ddd.ddd, where the number of digits after the decimal-point character is equal to the precision specification. If the precision is missing, it is taken as 6; if the precision is explicitly zero, no decimal- point character appears. If a decimal point appears, at least one digit appears before it.

12 is an int literal, 7.0 is a double literal. By the rules of expression evaluation int will be promoted to double and the result will be a double that is printed according to the format (f is not for float but for double).

If you want to print the value computed by a float division, you need to restrict to float using:

12/7.0f

The result will be float, and you may ask why do I use a double specifier then? Because, in any variadic function every float will be promoted to double. As printf is a variadic function... This is why there is no float specifier in format string of printf.

Upvotes: 2

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