C Language exercises

1. pointer exercise: swap the values of two pointer Variables

Level 1 pointer: * P, * q;

Int A = 0, B = 12; int * P = & A, * q = & B; printf ("before interchange: \ n % d, % d, % d \ n ", P, * P, & P); printf (" % d, % d, % d \ n ", Q, * q, & Q ); swap (p, q); printf ("after exchange: \ n % d, % d, % d \ n", P, * P, & P ); printf ("% d, % d, % d \ n", Q, * q, & Q );

Void swap (int * X, int * Y) {/* returns the same result using the XOR operation without intermediate variables */* x ^ = * Y; * y ^ = * X; * x ^ = * Y ;}

Output result:

Before exchange:

1606416820,0, 1606416808

1606416816,12, 1606416800

After exchange:

1606416820,12, 1606416808

1606416816,0, 1606416800

Exchange the addresses pointed to by two pointer variables:

Int A = 0, B = 12; int * P = & A, * q = & B; printf ("before interchange: \ n % d, % d, % d \ n ", P, * P, & P); printf (" % d, % d, % d \ n ", Q, * q, & Q ); swap (& P, * q); printf ("after exchange: \ n % d, % d, % d \ n", P, * P, & P ); printf ("% d, % d, % d \ n", Q, * q, & Q); void swap (int * X, int * Y) {/* the same result is obtained using the XOR operation without intermediate variables */* x ^ = * Y; * y ^ = * X; * x ^ = * Y ;}

Output result:

Before exchange:

1606416820,0, 1606416808

1606416816,12, 1606416800

After exchange:

1606416816,12, 1606416808

1606416820,0, 1606416800

Use a second-level pointer to exchange the addresses pointed to by two first-level pointer variables:

Int A = 2; int B = 3; int * P = & A; int * q = & B; int ** pp = & P; // defines the second-level pointer int ** QQ = & Q; // defines the second-level pointer printf ("before exchange: \ n % d, % d, % d \ n", P, * P, & P); printf ("% d, % d, % d \ n", Q, * q, & Q); swap (PP, QQ ); printf ("after exchange: \ n % d, % d, % d \ n", P, * P, & P); printf ("% d, % d, % d \ n ", Q, * q, & Q );

void swap(int **x,int **y){    int * temp;    temp=*x;    *x=*y;    *y=temp;}

Output result:

Before exchange:

1606416812,2, 1606416800

1606416808, 3, 1606416792

After exchange:

1606416808, 3, 1606416800

1606416812,2, 1606416792

2. Print the 9-9 multiplication table

void jiujiu(){    for(int i=1;i<10;i++){        for(int j=1;j<i+1;j++){                       printf("%d*%d=%d\t",j,i,i*j);                    }        printf("\n");    }}

Output result:

1*1 = 1

1*2 = 2
2*2 = 4

1*3 = 3
2*3 = 6 3*3 = 9

1*4 = 4
2*4 = 8 3*4 = 12
4*4 = 16

1*5 = 5
2*5 = 10 3*5 = 15
4*5 = 20 5*5 = 25

1*6 = 6
2*6 = 12 3*6 = 18
4*6 = 24 5*6 = 30
6*6 = 36

1*7 = 7
2*7 = 14 3*7 = 21
4*7 = 28 5*7 = 35
6*7 = 42 7*7 = 49

1*8 = 8
2*8 = 16 3*8 = 24
4*8 = 32 5*8 = 40
6*8 = 48 7*8 = 56
8*8 = 64

1*9 = 9
2*9 = 18 3*9 = 27
4*9 = 36 5*9 = 45
6*9 = 54 7*9 = 63
8*9 = 72 9*9 = 81

3. Input two integers at will. Use the two methods to output the largest one, and use the conditional statements and arithmetic operations to obtain the maximum value;

// Method 1: Determine the statement void bijiao1 () {int A, B; printf ("Please input two integers A, B :"); scanf ("% d", & A, & B); int max = A> B? A: B; printf ("maximum value: % d \ n", max);} // Method 2: arithmetic operation void bijiao2 () {int A, B; printf ("enter two integers A, B:"); scanf ("% d", & A, & B); int max = (A + B) + ABS (a-B)/2; // if you want to obtain the minimum value, write: int min = (a + B)-ABS (a-B )) /2; printf ("maximum value: % d \ n", max );}

Output result:

Please input two integersA, B:1 2

The maximum value is:2

Please input two integersA, B:3 4

The maximum value is:4

4. Print a heart shape:

    printf("  * *   * *\n");    printf(" *    *    *\n");    printf(" *         * \n");    printf("  *       *\n");    printf("   *     *\n");    printf("    *   *\n");    printf("      *\n");

5,Print out Yang Hui triangle

# Include <stdio. h> int main (INT argc, const char * argv []) {int A [10] [10]; // assign the sum of all the first and last bits to 1 for (INT I = 0; I <10; I ++) {A [I] [0] = 1; A [I] [I] = 1 ;} // A [I] [J] is the top two for (INT I = 2; I <10; I ++) {for (Int J = 1; j <I; j ++) {A [I] [J] = A [I-1] [J-1] + A [I-1] [J];} // print out the Yang Hui triangle for (INT I = 0; I <10; I ++) {// output the leading space for (int s = I; S <10; s ++) {printf ("") ;}for (Int J = 0; j <= I; j ++) {printf ("% d ", A [I] [J]);} printf ("\ n");} return 0 ;}

Output result:

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1

1 8 28 56 70 56 28 8 1

1 9 36 84 126 126 84 36 9 1