Why does gcc bitch about incompatible pointer types?
A variable int number;
A function func(int *number);
And a function call like this func(&number);
Compiles fine, and works fine. But when I change the type, for example to uint16_t number; func(uint16_t *number); gcc throws a warning about incompatible pointer type.
Does gcc hate me?
Name:
Anonymous2009-06-09 20:04
It's just you:
#include <iostream>
#include <stdint.h>
using namespace std;
void foo(uint16_t* pf)
{
cout << *pf << endl;
}
int main() {
uint16_t foobar = 42;
foo(&foobar);
}
NPP_SAVE: C:\Documents and Settings\James\Desktop\test.cpp
SET: Compiler = run@.cpp
$(COMPILER) = run@.cpp
NPP_EXEC: "run@.cpp"
g++ "C:\Documents and Settings\James\Desktop\test.cpp" -std=c++0x -o "C:\Documents and Settings\James\Desktop\test.exe"
Process started >>>
<<< Process finished.
"C:\Documents and Settings\James\Desktop\test.exe"
Process started >>>
42
<<< Process finished.
Name:
Anonymous2009-06-09 20:07
we all do
Name:
Anonymous2009-06-09 20:32
This thread is now about C++0x Variadic templates (type-safe printf!)
#include <iostream>
#include <vector>
#include <stdint.h>
#include <stdio.h>
using namespace std;
>>6
I don't get why you find this so bad. It may not be ``pretty'', but it does actually solve a real problem that needed to be solved, and it solves it adequately. Considering it's Sepples, it could be far, far worse.
>>7
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