CODEWITHSUNDEEP
c++c
aib
aib

Reputation: 46961

Determining the type of an expression

Sometimes I need to learn the type of an expression while programming in C or C++. Sometimes there's a good IDE or existent documentation to help me, but sometimes not. I often feel such a construct could be useful:

void (*myFunc)(int);
printf("%s", nameoftype(myFunc)); //"void (*)(int)"
int i, unsigned int u;
printf("%s", nameoftype(i+u));    //"unsigned int"

This is especially true for C++; think accessors of const objects - do they return a const reference or a copy? Think dynamic casts and templated classes.

How can I do this? (i.e. learn the type of an expression)

I use GCC but as far as I know, it does not have such an extension. So I guess I'm curious as to how people solve this problem. (Both compile-time and runtime solutions welcome.)

Upvotes: 8

Views: 7134

Answers (5)

Johannes Schaub - litb
Johannes Schaub - litb

Reputation: 507045

What are you looking for? Automatic type inference or looking for the type so you can declare a variable correctly manually? (your own answers look like you want to have the second one). In this case, consider using Geordi:

<litb> make type pointer to function taking pointer to array of 10 int returning void
<geordi> void (*)(int (*)[10])

<litb> geordi: { int a = -1; unsigned int b = 0; cout << ETYPE(a + b), ETYPE_DESC(a + b), (a + b); }
<geordi> rvalue unsigned int, rvalue unsigned integer, 4294967295

<litb> geordi: << TYPE_DESC(void (*)(int (*)[10]))
<geordi> pointer to a function taking a pointer to an array of 10 integers and returning nothing

Automatic type inference is not currently possible without helper libraries like boost.typeof, which will use compiler extensions like __typeof__ for GCC. Next C++ will get auto (with different semantics than current auto) and will be able to do that, together with decltype to get the type of an expression.

If you can live with getting out of local context, you can always create a function template like this:

template<typename T> void f(T t) { /* ... */ }
int main() { int a = -1; unsigned int b = 0; f(a + b); }

Upvotes: 2

Eyal
Eyal

Reputation: 5848

gcc has typeof() at compile time. It works like sizeof().

http://gcc.gnu.org/onlinedocs/gcc/Typeof.html has more information.

Upvotes: 1

aib
aib

Reputation: 46961

C++ has a typeid operator;

typeid(expression).name()

would return an implementation-defined name of the type of the expression. Alas, it is usually not human-readable.

Upvotes: 3

Anton Gogolev
Anton Gogolev

Reputation: 115769

Try Boost.Typeof to see if it fits.

Upvotes: 1

aib
aib

Reputation: 46961

Sometimes I just do:

int ***a = expression;

and look for the "<expression type> cannot be assigned to pointer-to^3 int" error. This seems to be the most portable workaround.

Upvotes: 21

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