C / C++ / MFC

Mircea Neacsu wrote:

IntelliSense will obligingly tell me that the compiler thinks it the variable type.

unordered_map<string, int> um{ {"foo", 1}, {"goo", 2}, {"boo",42}};

for (const auto & p : um)
{
  cout << p.first << ", " << p.second << "\n";
}

for (const pair<string, int> & p : um)
{
  cout << p.first << ", " << p.second << "\n";
}

Maximilien wrote:

when I declare a variable with auto, it will be typed accordingly and I cannot change the type.

char* b = "ABCD"; // means treat b as a char pointer

int* a = (int*)b; // means please treat b as a int pointer just this one time

int somefunction(int* pi, int count)
{
    int sum = 0;
    for (int i = 0; i < count; ++i)
    {
        sum += *pi;
    }
    return sum;
}

// and you call it like this

   char* ci = "Foobar";
   int total = somefunction((int*)ci, strlen(ci));

k5054 wrote:

if the target CPU ...

Richard MacCutchan wrote:

If you have something like the following:

Richard MacCutchan wrote:

When you use a cast the compiler does what can be done at machine level, but the object itself is still a char*, and any attempt to use it in any other way will raise a compiler error.

Richard MacCutchan wrote:

And the result of calling that function may, or may not, make sense.

trønderen wrote:

while in other cases (other interpretations), it will create a new bit pattern

for(const auto& x: my_container)
{
    // ...
}

std::vector<short> v;
// ...
short n = v[0];

std::vector<int> v;
// ...
short n1 = v[0];                    // oops, variable type should be changed as well
auto n2 = v[0];                     // this is OK
decltype(v)::value_type n3 = v[0];  // this is also OK, though I don't like it