C++ convert a float array to float* (pointer)

Question

I wonder how to convert a float array to a float* I have this situation :

float* floatTab = {12f, 0.5f, 3f};

It gives me an error here. but if I write it like this float floatTab[3] = {12f, 0.5f, 3f};it compiles alright.

This sounds like an XY problem. What are you trying to solve? — juanchopanza, Nov 17 '14 at 11:07
I have a framework that uses a float* i can't pass a float[] so i can't use the code that compiles — Driss Bounouar, Nov 17 '14 at 11:08
floatTab will *decay* to a float* if used as a function parameter, but use the code that "complile [sic] alright" to create the array. — Bathsheba, Nov 17 '14 at 11:09
I have this error when trying to compile the above code excess elements in scalar initializer — Driss Bounouar, Nov 17 '14 at 11:32
@DrissBounouar A pointer is not an array. That is the simple explanation for the error. That initialization is for arrays, not a pointer. A pointer is a single, integral value, not a set of values. — PaulMcKenzie, Nov 17 '14 at 11:45

score 6 · Accepted Answer · answered Nov 17 '14 at 11:09

6

This works OK:

float floatTab[3] = {12f, 0.5f, 3f}; float* ptr = floatTab;

answered Nov 17 '14 at 11:09

FunkyCat

Thanks. I was doing some extra changes to the same above code trying to give a reference instead of doing as simple as this. – Driss Bounouar Nov 17 '14 at 12:45

vincentp · Answer 2 · 2014-11-17T11:40:10.963

1

Prefer STL containers instead of C arrays (or others RAII-conform classes):

const std::array<float, 3> array = { 1.f, 2.f, 3.f };
float *ptr = &array[0];

Don't forget to include <array> and <initializer_list> to compile this code.

edited Nov 17 '14 at 11:40

answered Nov 17 '14 at 11:19

vincentp

2 Answers2