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Details of C that every not so noob forgets


I already wrote a post about Details of C that every noob forgets. In this post I will write about more advanced details. I hope this will improve your understand of the obscure C. By the way, for the a complete review I recommend to read Expert C Programming: Deep C Secrets by Peter van der Linden or my notes.

WTF? const is not constant

The keyword const doesn’t mean constant expression , it only gives the status of read-only.

const int two=2;

switch (i) {
    /*case const-expr: statements*/
     case 1: printf("case 1 \n");
     case two: printf("case 2 \n");
**error** ^^^ integral constant expression expected
     case 3: printf("case 3 \n");
     default: ; }

Typedef vs macros

Even though they might look the same, these guys are totally different.

#define type1 int

typedef  int type2;

int main(){

    type1 a; // works
    type2 b; // works

    unsigned type1 c; // works
    unsigned type2 d; // illegal
}

typedef defines a new data type, while #define is just a replacement. For example typedef int type2; means that type2 is int, whereas #define type1 int means that type1 will be replaced by int every time it appears in the code. Thus, you can even to the following:

#define type1 int
typedef  type1 type2;

Again #define does not define a new data type.

#define int_ptr int *
int_ptr a,b; 
/*
 this means:
  int *a, b;
 which is different from
  int *a, *b;
*/

Do not self-deSTRUCT

I used to make this mistake when I was learning C. Be careful when using typedef with struct.

// struct with foo as a tag
struct foo{
    int a;
    char b;
};

// foo1 is a struct foo
typedef struct foo{
    int a;
    char b;
} foo1;

Check the following statements:

foo a;  // illegal
// there is not 'foo' type 
struct foo a; // good
foo1 a; // good
struct foo1 a; // illegal
// foo1 is already a data type

EPIC BATTLE: arrays vs pointers

Accessing data from array is different if we use pointers.

char a[5] = "abcd";
c = a[2];

/*
in symtab: a is on 0x4000 
a = | a | b | c | d | '\0' |
*/

If we want to access the third element a[2]

  1. 0x4000 + 2 , directly!!
  2. get value from 0x4002
0x0000000000400509 main+19 movzbl -0x1e(%rbp),%eax
0x000000000040050d main+23 mov    %al,-0x1(%rbp)

In case of pointers:

char *p = a;
c = *(p+2);

To access the third element:

  1. get address of p
  2. p+2
  3. get value from that address
0x000000000040052d main+55 mov    -0x10(%rbp),%rax
0x0000000000400531 main+59 movzbl 0x2(%rax),%eax
0x0000000000400535 main+63 mov    %al,-0x1(%rbp)

The array definition will allocate N bytes, whereas a pointer definition will only allocate a word size.

int m[3];
/* sizeof(m) = 12 */
int *p;
/* sizeof(p) = 8 (CPU of 64 bits)*/

Another example:

char a[] = "hallo";
char *b  = "hallo";

a[] is an array that holds {'h','a','l','l','o','\0'}, and it can be modified. *b is a pointer to a read-only section that holds "hallo".

One last example:

int main(){
    // an array 
    char a[] = "hola";

    // a pointer 
    char *b;

    // a pointer to read-only memory
    char *c = "hallo";

    // b can point to a or c
    b = c;
    b = a;

    // c can point to a or c
    c = a;
    c = b;

    // but a cannot point to a or c, 
    // because it is not a FUCKING POINTER!!!!!

    a = b;
   // error: assignment to expression with array type   a = b;

}

When an array is a pointer

Declarations themselves can be further split into three cases:

  • declaration of an external array extern char a[];. a cannot be rewritten as a pointer.
  • definition of an array (remember, a definition is just a special case of a declaration; it allocates space and may provide an initial value). char a[10];. a cannot be rewritten as a pointer.
  • declaration of a function parameter int function ( char a[]); . a is rewritten as a pointer whether you like it or not.
void my_function_1( int a [2][3][5] ) { ; }  
void my_function_2( int a [][3][5] ) { ; }  
void my_function_3( int (*a)[3][5] ) { ; }  
 

int main(){
    int arr[2][3][5];  

    my_function_1( arr );  
    my_function_2( arr );  
    my_function_3( arr );  

    int (*p) [3][5] = arr;  

    my_function_1( p );  
    my_function_2( p );  
    my_function_3( p );

    int (*q)[2][3][5] = &arr;

    my_function_1( *q );
    my_function_2( *q );
    my_function_3( *q );
}

In summary, the rule array name is rewritten as a pointer argument isn’t recursive. An array of arrays is rewritten as a pointer to arrays NOT as a pointer to pointer.

Thus,

  • array of array char a[9][4] can be rewritten as char (*a)[4] pointer to array
  • array of pointers char *a[4] can be rewritten as char **a pointer to pointer
  • pointer to array char (*a)[4] is char (*a)[4] pointer to array
  • pointer to pointer char **a is char **a pointer to pointer

The very common case is the main function:

int main(int argc, char **argv){}

int main(int argc, char *argv[]){}

Just for fun

Read the following one-liner:

main() { printf(&unix["\021%six\012\0"],(unix)["have"]+"fun"-0x60);}`

David Korn wrote this one-liner and won the iocc contest in 1987

Try to understand yourself the code, there are some extra info you might need:

  • unix is a predefined macro with value 1 #define unix 1
  • a[i] is resolved by the compiler as *(a+i)
  • a[i][j] is resolved by the compiler as *( *(a+i) + j)

Last words

C is a very simple language if you compare to python or even c++, and still very powerful. I hope you enjoy all the information provided here. Let me know if you have some questions.

Have fun!


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