Using a typedef'd uint causes error, while “unsigned int” does not…?

问题

For some reason, when I define a variable as "uint" instead of "unsigned int" in my program, it errors. This seems strange, because uint is typedef'd as:

typedef unsigned int uint;

...so I would think that I could use the two interchangeably. To be more exact, I am assigning the result of a function which returns "unsigned int" into a uint variable, then using that uint in a vector resize call... at which point it errors. Ie, my code looks something like this:

unsigned int getUInt()
{
    return 3;
}

int main(void) {
    vector<vector<float> > vectVect(100000);
    for(uint i = 0; i < vectVect.size(); ++i)
    {
        vector<float>& myVect = vectVect[i];
        uint myUnsignedInt = getUInt();
        myVect.resize(myUnsignedInt);
    }
    cout << "finished" << endl;
}

...and the line it errors at is the myVect.resize line.

Obviously, I already have a solution, but I'd like to understand WHY this is happening, as I'm pretty baffled. Anyone have any ideas?

PS - In case anyone thinks it may matter, I'm using gcc v4.1.2 on fedora 15... and the include file which defines uint is /usr/include/sys/types.h.

回答1:

My guess is that there is another uint in the system. Try renaming yours to something unusual or even better wrap it in a namespace.

namespace MY {
    typedef unsigned int uint;
}

for (MY::uint i = 0; ....

回答2:

My guess is that it's trying to form a "hierarchy" of some sort.

In other words, we have:

typedef unsigned int size_t;
typedef unsigned int uint;

If size_t is "more specific" than a generic unsigned int, then it makes sense to prevent it from being converted to uint, which may be "more specific" than any old unsigned int.

I would expect this to be a warning if anything, though, not an error...

来源：https://stackoverflow.com/questions/8163262/using-a-typedefd-uint-causes-error-while-unsigned-int-does-not

标签

c++

gcc

typedef

unsigned

uint