Explicit type casting operator in C/C++

Question

The following code converts the float type 7.5 to an integer value 7, the remainder is lost. Here, the typecasting operator is int. I know it is valid typecast in C++.

int main()
{
        int i;
        float f = 7.5;
        i = (int) f; // Valid in C/C++
}

But another way to do the same thing in C/C++ is to use the functional notation preceding the expression to be converted by the type and enclosing the expression between parentheses:

i = int (f); // It is worked in C++ but not C

So, I have a question, Is it valid way to typecast in C++?

also take a look here: http://stackoverflow.com/questions/332030/when-should-static-cast-dynamic-cast-const-cast-and-reinterpret-cast-be-used — xtofl, Oct 06 '16 at 05:37
there's no C/C++ language. It's one of C++ typecast syntaxes so read a book before asking things like this http://en.cppreference.com/w/cpp/language/explicit_cast http://stackoverflow.com/q/7558837/995714 — phuclv, Oct 06 '16 at 05:41
@R Sahu Sorry, It is not work in C but In C++, it working fine. — msc, Oct 06 '16 at 05:52

xtofl · Answer 1 · 2016-10-07T05:38:17.953

2

It is in C++. But although it is valid, C++ (opposed to C) encourages explicitness in your casting:

auto i = static_cast<int>(7.0);

(Oh, and use the type safe streams instead of the error-prone printf:

std::cout << i << "\n";

edited Oct 07 '16 at 05:38

answered Oct 06 '16 at 05:34

xtofl

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Good answer. I'd mention the type safety improvement by using static_cast. – Ely Oct 06 '16 at 05:35
It is valid in C++ but –unless I am mistaken – not valid in C. – Martin R Oct 06 '16 at 05:40
@MartinR, you are correct. It is not valid in C. See [my answer](http://stackoverflow.com/a/39888307/434551). – R Sahu Oct 06 '16 at 05:48
@MartinR: thanks. Indeed - I know little of C. Corrected – xtofl Oct 07 '16 at 05:38

R Sahu · Accepted Answer · 2016-10-06T05:47:14.147

i = int (f);

is valid in C++ but not in C.

From the C99 Standard, 6.5.4 Cast operators

 cast-expression:
      unary-expression
      ( type-name ) cast-expression

C++ supports the above form of casting as well as function style casting. Function style casting is like calling the constructor of a type to construct on object.

Check out the following block code compiled using C and C++.

#include <stdio.h>

int main()
{
   float f = 10.2;
   int i = int(f);

   printf("i: %d\n", i);
}

Non-working C program: http://ideone.com/FfNR5r

Working C++ program: http://ideone.com/bSP7sL

Explicit type casting operator in C/C++

2 Answers2

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