C-Kurs/(Pointer) Swapping/Musterlösung

2010-09-17T03:42:29Z

92.206.6.50:

1. Teilaufgabe

#include <stdio.h>

void swap(int *x, int *y) {

int temp= *x;
*x= *y;
*y= temp;
}

int main(){

int i= 1;
int j= 2;

printf("i: %i j: %i\n", i, j);

swap(&i, &j);

printf("i: %i j: %i\n", i, j);

return 0;
}

2. Teilaufgabe

#include <stdio.h>

void swap(int *x, int *y) {

int temp= *x;
*x= *y;
*y= temp;
}

void print_int_array(char* name, int* array, int size){

int i;

printf("%s:", name);
for(i= 0; i<size; i++){
printf(" %i", array[i]);
}
printf("\n");
}

int main(){

int a[]= {1, 2, 3, 4};
int b[]= {5, 6, 7, 8};

int length= sizeof(a) / sizeof(int);

print_int_array("a", a, length);
print_int_array("b", b, length);

int i;
for(i=0; i<length; i++){
swap(&a[i], &b[i]);
}

print_int_array("a", a, length);
print_int_array("b", b, length);

return 0;
}

3.Teilaufgabe

#include <stdio.h>

void swap(int **x, int **y) {

int* temp= *x;
*x= *y;
*y= temp;
}

int main(){

int i= 1;
int j= 2;

int* p1= &i;
int* p2= &j;

printf("------------------------------------------\n");
printf("before swap:\n");
printf("------------------------------------------\n");
printf(" p1:\t%p\t p2:\t%p\n", p1, p2);
printf("*p1:\t%10i\t*p2:\t%10i\n", *p1, *p2);
printf(" i:\t%10i\t j:\t%10i\n", i, j);

swap(&p1, &p2);

printf("------------------------------------------\n");
printf("after swap:\n");
printf("------------------------------------------\n");
printf(" p1:\t%p\t p2:\t%p\n", p1, p2);
printf("*p1:\t%10i\t*p2:\t%10i\n", *p1, *p2);
printf(" i:\t%10i\t j:\t%10i\n", i, j);

return 0;
}

4.Teilaufgabe

#include <stdio.h>

void swap(int **x, int **y) {

int* temp= *x;
*x= *y;
*y= temp;
}

void print_int_array(char* name, int* array, int size){

int i;

printf("%s:", name);
for(i= 0; i<size; i++){
printf(" %i", array[i]);
}
printf("\n");
}

int main(){

int a[]= {1, 2, 3, 4};
int b[]= {5, 6, 7, 8};

int length= sizeof(a) / sizeof(int);

// a and b are constant
// alternatively, you can use calloc

int *front= &a[0];
int *back= &b[0];

print_int_array("front", front, length);
print_int_array("back ", back, length);

swap(&front, &back);

print_int_array("front", front, length);
print_int_array("back ", back, length);

return 0;
}

5. Teilaufgabe

#include <stdio.h>

void swap(void **x, void **y) {

void *temp = *x;
*x = *y;
*y = temp;
}

void print_int_array(char* name, int* array, int size){

int i;

printf("%s:", name);
for(i= 0; i<size; i++){
printf(" %i", array[i]);
}
printf("\n");
}

void print_float_array(char* name, float* array, int size){

int i;

printf("%s:", name);
for(i= 0; i<size; i++){
printf(" %f", array[i]);
}
printf("\n");
}

int main(){

int i= 1;
int j= 2;

int* p1= &i;
int* p2= &j;

printf("i: %i j: %i\n", *p1, *p2);

swap((void*)&p1, (void*)&p2);

printf("i: %i j: %i\n", *p1, *p2);

int a[]= {1, 2, 3, 4};
int b[]= {5, 6, 7, 8};

int length= sizeof(a) / sizeof(int);

// a and b are constant
// alternatively, you can use calloc

p1= &a[0];
p2= &b[0];

print_int_array("a", p1, length);
print_int_array("b", p2, length);

swap((void**)&p1, (void**)&p2);

print_int_array("a", p1, length);
print_int_array("b", p2, length);

float c[]= {1.0, 2.0, 3.0, 4.0};
float d[]= {5.0, 6.0, 7.0, 8.0};

length= sizeof(c) / sizeof(int);

// c and d are constant
// alternatively, you can use calloc

float *p3= &c[0];
float *p4= &d[0];

print_float_array("x", p3, length);
print_float_array("y", p4, length);

swap((void**)&p3, (void**)&p4);

print_float_array("x", p3, length);
print_float_array("y", p4, length);

return 0;
}

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C-Kurs/(Pointer) Swapping/Musterlösung