GCC warning about assigning a pointer variable with an array?

Question

code snippet:

int *c[2] = {{1,2,3}, {4,5,6}};

gives warning:

warning: incompatible integer to pointer conversion initializing 'int *' with an expression of type 'int'
      [-Wint-conversion]
    int *c[2] = {{1,2,3}, {4,5,6}};
                  ^
warning: excess elements in scalar initializer
    int *c[2] = {{1,2,3}, {4,5,6}};
                    ^

I suppose array {1,2,3} would decay to pointer so the assignment be legit?

Further more, according to the warning, why does the compiler think i'm trying to assign int to int *? instead of int array type to int *? Thanks in advance!

You are trying to initialise int c[2][3] with the given data. — Weather Vane
– Weather Vane, Commented Apr 24, 2019 at 18:04
Hi Weather, c is a 2-element array of pointers, you can gcc -S to check the assembly implementation compared with int cc[2][3]. — mzoz
– mzoz, Commented Apr 24, 2019 at 18:07

dbush · Accepted Answer · 2019-04-24 18:14:07Z

3

Bracketed initializers are not arrays, and so cannot undergo decay to a pointer type. And because c is an array of int *, each initializer needs to be of that type. Nested initializers only work with actual arrays (not pointers) or structs.

What you can do however is use compound literals in the initializer which do have an array type.

int *c[2] = {(int []){1,2,3}, (int []){4,5,6}};

edited Apr 24, 2019 at 18:14

answered Apr 24, 2019 at 18:04

dbush

233k27 gold badges261 silver badges334 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

mzoz Over a year ago

Thanks dbush, but why does compiler say the code is initializing int * with int? Does initializer have the type of int?

dbush Over a year ago

@mzoz Each value in the initializer list is an int, so that's what it attempts to assign to the elements of c. The extra braces don't make the initializer an array.

John Bollinger · Accepted Answer · 2019-04-24 18:07:05Z

1

I suppose array {1,2,3} would decay to pointer so the assignment be legit?

No. {1, 2, 3} is not an array, and no decay applies to it. It is an initializer suitable for initializing an array of three or more elements of arithmetic type, but you are trying to use it to initialize a pointer. You could do something like this, instead:

static int x[] = { 1, 2, 3 };
static int y[] = { 4, 5, 6 };
int *c[] = { x, y };

Or you could use compound literals to avoid declaring variables x and y, as another answer suggests. In either of these cases, the initializer elements are arrays, and they do decay to pointers.

answered Apr 24, 2019 at 18:07

John Bollinger

191k11 gold badges103 silver badges206 bronze badges

2 Comments

mzoz Over a year ago

Thanks John, but why does compiler say the code is initializing int * with int? Does initializer have the type of int?

John Bollinger Over a year ago

The first element of {1, 2, 3} does, and that's what the compiler is looking at. The initializer for a scalar (including a pointer) can optionally be enclosed in {}. GCC has diagnosed a small-picture item instead of recognizing and diagnosing the larger issue, and it is understandable that it does so.

Collectives™ on Stack Overflow

GCC warning about assigning a pointer variable with an array?

2 Answers 2

2 Comments

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related