[C ++ Basics] Details few people know about C ++

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• 3. Macro-defined operations
• 4. sizeof (union)
• 5. Array Memory Allocation
• 6. Char <=> int

1. Unsigned int automatic conversion

Void foo () {unsigned int a = 6; int B =-20; (a + B> 6 )? Puts ("> 6"): puts ("<= 6"); // output> 6/* test whether you understand the principle of automatic Integer Conversion in C, I found that some developers know little about these things. In any case, the answer to this unsigned integer is "> 6 ". The reason is that when there are signed and unsigned types in the expression, all operands are automatically converted to the unsigned type. Therefore,-20 is a very large positive integer, so the result calculated by this expression is greater than 6. This is very important for embedded systems that should frequently use the unsigned data type. */}

2. array community

Union {int I; char x [2];} a; void fun () {. x [0] = 10;. x [1] = 2; printf ("% d-% d \ n",. i,. x [0],. x [1]); // 522-10-2: Description. x [0] and. x [1] has no influence on each other, so their addresses are different. A. I =. x [0] +. x [1] * 256 (Power 8 of 2);. I = 1027; printf ("% d-% d \ n",. i,. x [0],. x [1]); // 1027-3-4 1027 = (4) * 256 + (3). x [0] = 300;. x [1] = 10; printf ("% d-% d \ n",. i,. x [0],. x [1]); // 2604-44-10,. x [0] = 300-256 = 44;. I =. x [0] +. x [1] * 256;}/* union is a space used by internal variables. allocated according to the size, int I occupies 4 bytes, char x [2] occupies 2, so a total of 4 bytes are allocated. A total of 4 bytes of memory, corresponding to x is equivalent to taking up 2 lower bytes, and 1, and 10 assigned to x, there are two and ten digits (hexadecimal) the public body shares the size of the sizeof (a) in a memory area that is the longest member in the shared body. That is, I int: (| _ |) (| _ |) (| _ |) char x [2]: (| _ |) ^ high address and low address analysis:. x [0] = 10 ======================================== ===================> (| 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 |) a. x [1] = 2 ====================================== ========> (| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 |) if the length of the Public sizeof (int) is 4-byte 32, the shared memory is: (| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |) (| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |) (| 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 |) (| 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 |). if I is a 4-byte integer, I = 2 ^ 9 + 2 ^ 3 + 2 ^ 1 = 256 + 8 + 2 = 522 */

3. Macro-defined operations

#include <iostream>using namespace std;#define SQR(X) X*Xint main(void){int i=10,j=5,n=10;n*=i+j;//n=n*(i+j)cout<<n<<endl;//150int a = 10,k = 2, m=1; a /= SQR(k+m)/SQR(k+m); //a/=(k+m*k+m/k+m*k+m)cout<<a<<endl;//1return 0;}

4. sizeof (union)

#include <iostream>using namespace std;union a {int a_int1;double a_double;int a_int2;};typedef struct{a a1;char y;} b;class c{double c_double;b b1;a a2;};int main(void){cout<<sizeof(a)<<endl;//8cout<<sizeof(b)<<endl;//16cout<<sizeof(c)<<endl;//32return 0;}

5. Array Memory Allocation

Char str1 [] = "abc"; char str2 [] = "abc"; const char str3 [] = "abc"; const char str4 [] = "abc "; const char * str5 = "abc"; const char * str6 = "abc"; char * str7 = "abc"; char * str8 = "abc "; cout <(str1 = str2) <endl; cout <(str3 = str4) <endl; cout <(str5 = str6) <endl; cout <(str7 = str8) <endl; Result: 0 0 1 answer: str1, str2, str3, str4 are array variables, they have their own memory space; str5, str6, str7, and str8 are pointers pointing to the same constant area.

6. char <=> int

// One integer to Char char x = 5 + '0 '; // '5' // a char to int y = '5'-'0 '; // 5 // obtain the 'D' char z = 'A' + 3 after 'a'; // 'D'