typedef int pipe_t[2]; 是什么意思？

Question

Can anyone explain to me in a very simple way the meaning of these lines of code.

typedef int pipe_t[2];
pipe_t *piped; 
int L; 
L = atoi(argv[2]);
piped = (pipe_t *) malloc (L*sizeof(pipe_t));

Answer 1

• Type pipe_t is "array of 2 ints"
• Variable piped is a pointer to such arrays.
• L is an integer, assigned from the command line
• Pointer piped is assigned to point to a block of memory large enough for L arrays of the type above.

Answer 2

For this typedef, you can read the declaration from right to left, so in

typedef int pipe_t[2];

you have

• [2] says it is an array of 2. Positions [0] and [1]
• pipe_t is the variable (type) name
• int says pipe_t[2] is an array of 2 int
• typedef says that it is, in fact, a type — an alias for a user-defined type representing an array of 2 int .

Run this program

#include<stdio.h>

int main(int argc, char** argv)
{
    typedef int pipe_t[2];
    printf("{typedef int pipe_t[2]} sizeof(pipe)_t is %d\n",
        sizeof(pipe_t));

    pipe_t test;

    test[1] = 2;
    test[0] = 1;
    printf("pair: [%d,%d]\n", test[0], test[1]);

    // with no typedef...
    int    (*another)[2];
    another = &test;
    (*another[0]) = 3;
    (*another)[1] = 4;
    printf("{another} pair: [%d,%d]\n", (*another)[0], (*another)[1]);

    pipe_t* piped= &test;
    printf("{Using pointer} pair: [%d,%d]\n", (*piped)[0], (*piped)[1]);
    return 0;
};

And you see

{typedef int pipe_t[2]} sizeof(pipe)_t is 8
pair: [1,2]
{another} pair: [3,4]
{Using pointer} pair: [3,4]

And you see that pipe_t has a size of 8 bytes, corresponding to 2 int 's in x86 mode. And you can declare test as pipe_t and use it as an array of int . And the pointer works the same way

And I added the code to be used without such typedef, so we see that using a typedef makes it a bit cleaner to read.