Drop ampersands from that line. The problem is that the & operator in this context produces an expression with a different type than what FFT expects. FFT expects a pointer to a double and &werte produces a pointer to an array of N elements. So, in order to make FFT happy, just pass werte which … Read more
You can’t return arrays from functions in C. You also can’t (shouldn’t) do this: returned is created with automatic storage duration and references to it will become invalid once it leaves its declaring scope, i.e., when the function returns. You will need to dynamically allocate the memory inside of the function or fill a preallocated buffer … Read more
The primary reason, you cannot do bitwise operation on a void *, but you can do the same on a intptr_t. On many occassion, where you need to perform bitwise operation on an address, you can use intptr_t. However, for bitwise operations, best approach is to use the unsigned counterpart, uintptr_t. As mentioned in the … Read more
Use the format specifier %p: The standard requires that the argument is of type void* for %p specifier. Since, printf is a variadic function, there’s no implicit conversion to void * from T * which would happen implicitly for any non-variadic functions in C. Hence, the cast is required. To quote the standard: 7.21.6 Formatted … Read more
A zero-size vector of pointers: A vector of NULL pointers: A vector of pointers to newly allocated objects (not really initialization though): Initializing a vector of pointers to newly allocated objects (needs C++11): A smarter version of #3:
typedef is a language construct that associates a name to a type.You use it the same way you would use the original type, for instance using them like As you can see, you could just replace the typedefed name with its definition given above. The difficulty lies in the pointer to functions syntax and readability … Read more
&arr gives an array pointer, a special pointer type int(*)[5] which points at the array as whole. arr, when written in an expression such a int *q = arr;, “decays” into a pointer to the first element. Completely equivalent to int *q = &arr[0]; In the first case you try to assign a int(*)[5] to … Read more
It looks like the line is outside the main() function. You can’t do that, because you can’t execute code outside functions. The only thing you can do at global scope is declaring variables. Just move it inside the main() or any other function, and the problem should go away. (PS: Take a look at this … Read more
When you pass a pointer by a non-const reference, you are telling the compiler that you are going to modify that pointer’s value. Your code does not do that, but the compiler thinks that it does, or plans to do it in the future. To fix this error, either declare x constant or make a … Read more
This is because you have not allocate your objects prior to using them: It’s worth adding that using pointers in this situation is, well, pointless: string objects in the standard C++ library allocate the data for the string from the heap; strings are usually not much more than a pair of pointers anyway.