int foo[5]; What is the type of foo? You should know this.

foo is a pointer to an int array of size 5.

>>60086901
in C

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264KB, 1280x853px

>>60086901

>foo is a pointer
Whoops! Would you like to try again?

>>60086993
*foo is equivalent to foo[0]
*(foo + 4) is equivalent to foo[4]

so yes, it's a pointer.

>>60086855
"array of 5 integers"
What is the point of this thread?

>>60087041
arrays are convertibles to pointers through a standard conversion but aren't pointers.

>>60087063
>Can be dereferenced like pointers

FTFY

>>60086993
foo is an array of 5 ints.

So, of course, &foo is a pointer to an array of 5 ints.

Try running

printf("%p %p %p %p", foo, foo + 1, &foo, &foo + 1);

>>60087109

int foo[5];
foo[0] = 2;
printf("%d %p %p", *foo, foo, &foo);

2 0x7ffecdb99b00 0x7ffecdb99b00

>>60087106

>I don't know what decay is

>>60086855
The type is "array of 5 int".

fcking retards
foo is pointer, array is only imaginary word
foo points to address based on [x] as foo + type * x

>>60087109
&foo is a reference
*fpointer = &foo is a pointer

>>60087235
>array is only imaginary word
wow, it's not even a real word? Look at that!

its type is
>sub esp, 5*sizeof(int)

>>60087250
Arrays are automatically passed as a ref went sent as a parameter to functions..

>>60087161
And what about

foo + 1

and

&foo + 1

? In this example they should end in 9b04 and 9b14 respectively.

>>60087293
Foo+1 4 bytes
Foo+2 10 bytes
M'kay

>>60087235

Use sizeof senpai, you'll be surprised

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24KB, 492x299px

Use vectors

>>60086855
Stop using C.

>>60087329

    printf("sizeof(int):    %d\n", sizeof(int));
    printf("sizeof(int*):   %d\n", sizeof(int *));
    printf("sizeof(int **): %d\n", sizeof(int **));
    
    int foo[5];
    foo[0] = 2;
    printf("%d %p %p %p %p\n", *foo, foo, &foo, foo+1, &foo+1);

2 0x7ffc80a982c0 0x7ffc80a982c0 0x7ffc80a982c4 0x7ffc80a982d4

>>60087442
fuck forgot the rest

sizeof(int):    4                                                                                                                                                                                               
sizeof(int*):   8                                                                                                                                                                                               
sizeof(int **): 8

>>60087456
I don't you did that correct

Wouldn't it make sense that a pointer is 64 bits on a 64bit arch? Int is only guaranteed to be between char and long

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47KB, 306x469px

>>60086855
>this thread
you really need to find more interesting things to argue about

Arrays and pointers are actually slightly different in C. In C an array also has information about the number of elements if it was allocated on the stack. This allows you to use sizeof(array)/sizeof(array[0]) to get the number of elements in the array. This only works on arrays and not pointers as the size of a pointer is just whatever size your machine uses (32 or 64 bits).

>>60086855
I know it's type is int[5], but only because I saw some of the retarded sycophants on #C on freenode pitch a fit about it once to satisfy one of that channel's admins.

>>60089101
*its
fucked that up, was originally going to type "it's int[5]"...