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Alessio1989

casting an expression: what exactly casts the compiler?

6 posts in this topic

If I cast an expression with C++ casts, or even C-style casts, what exactly will cast the compiler?

 

Does the compiler cast every symbol of the expression before evaluating it or does it cast only the final expression value without casting the expression symbols?

 

eg: if I want cast only the final result of the expression without changing the symbols type, do I need to cast every symbol of the expression of its own type and then cast the entire expression to the target type or I don't need to do that?

Edited by Alessio1989
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When you do the cast only the result of the expression contained within the cast is casted. However, the compiler automatically casts certain things within an expression. For example, if you multiply a float variable by an integer variable the integer will be casted to a float causing the operation to be float * float. I believe most/all of these conversions are well-defined by the standard, but can't say for sure. There might be some confusing unsigned/signed casting quirks that are unspecified or undefined.
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it should only cast the result afaik.

 

I hope not! It should cast whatever is directly in front of the cast operator. So, in this case:

[source]
int a=1, b = 2;

float f = (float)a/b;
[/source]
 

It will convert both a and b to floats prior to computing the division. The cast itself will only actually force 'a' to be converted to a float, but since 'a' is a float, 'b' will in turn be converted to a float (due to the rule that the operand on the left of the operator determining the operation used). As proof, the ASM looks like this:

 
[source]
cvtsi2ss xmm0, DWORD PTR _a$[ebp] // convert a to float
cvtsi2ss xmm1, DWORD PTR _b$[ebp] // convert b to float
divss xmm0, xmm1
movss DWORD PTR _c$[ebp], xmm0
[/source]
 
In this case however, it will evaluate a/b using integers, and then cast the result.

[source]
int a=1, b = 2;

float f = (float)(a / b);
[/source]
 
The ASM: 
 
[source]
mov eax, DWORD PTR _a$[ebp]
cdq
idiv DWORD PTR _b$[ebp] // integer division
cvtsi2ss xmm0, eax // convert result to float
movss DWORD PTR _d$[ebp], xmm0
[/source]
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due to the rule that the operand on the left of the operator determining the operation used


There is no such rule.

Each operand is equal in its consideration of which operator overload to invoke or which implicit conversion to perform. The following would also have an implicit cast to float:
 
int a; float b;
auto c = a / b; // decltype(c) ==> float
Implicit conversions have a (relatively) complex set of rules about what can be cast to what and when. Which side of an operator each operand on is not a part of those rules.

With user-defined types nothing stops you from overloading operators with operands in whatever order you choose, so long as at least one operand is a user-defined type:
 
struct A {};
A operator/(A l, float r);
A operator/(float l, A r);
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