So I'm new to C, I'm trying to make procedure to swap value of 2 variables.
When I run this the swap didn't work.
#include <stdio.h>
#include <stdlib.h>
void swap(int A,int B)
{
A = A + B;
B = A - B;
A = A - B;
}
int main(void){
int num1,num2;
printf("insert first number :\n");
scanf("%d",&num1);
printf("insert second number :\n");
scanf("%d",&num2);
swap(num1,num2);
printf("%d %d\n",num1,num2);
return 0;
}
That is because arguments are passed by value and modifying them in callee won't affect caller's local variables. Use pointers to modify ones.
#include <stdio.h>
#include <stdlib.h>
void swap(int* A,int* B)
{
*A = *A + *B;
*B = *A - *B;
*A = *A - *B;
}
int main(void){
int num1=0,num2=0; /* initialize for in case scanf() fails */
printf("insert first number :\n");
scanf("%d",&num1);
printf("insert second number :\n");
scanf("%d",&num2);
swap(&num1,&num2);
printf("%d %d\n",num1,num2);
return 0;
}
It didn't work because when you call the procedure, it is made a copy of the arguments to the new scope.
The right way is with the address of the variables.
Change to:
swap(&num1,&num2);
and
void swap(int *A,int *B)
{
*A = *A + *B;
*B = *A - *B;
*A = *A - *B;
}
Related
A function receives two integer pointers, int* a and int* b. Set the value of *a to their sum, and *b to their absolute difference.
There is no return value, and no return statement is needed.
I got the values for *a but I'm unable to get the code for *b.
#include <stdio.h>
void update(int *a,int *b);
int main() {
int a, b;
int *pa = &a, *pb = &b;
scanf("%d %d", &a, &b);
update(pa, pb);
printf("%d\n%d", a, b);
return 0;
}
void update(int *a,int *b)
{
*a+=*b;
*b=*a-*b;
}
You may use a temporally variable to store value of *a and then use it:
int tmp = *a;
*a += *b;
*b = abs(tmp - *b);
A little cleanup. Only the function needs pointers. Your main does not.
#include <stdio.h>
#include <stdlib.h> // abs()
void sum_and_diff( int * a, int * b )
{
int sum = *a + *b;
int diff = abs( *a - *b );
*a = sum;
*b = diff;
}
int main(void)
{
int a, b;
printf( "a? " ); scanf( "%d", &a );
printf( "b? " ); scanf( "%d", &b );
sum_and_diff( &a, &b );
printf( "sum = %d\n", a );
printf( "absolute difference = %d\n", b );
return 0;
}
A simple answer (the simplest?), that doesn't require temporary variables or external functions like abs(), is:
void update(int *a, int *b)
{
*a += *b;
*b = *a-*b-*b;
}
It fixes your code by also subtracting the "extra" *b you just added to *a
I m learning c (programming in ANSI c -> fifth edition) and facing the below issue:
I'm implementing one program with pointers to functions
#include<stdio.h>
//void swap (int *a, int *b); //function declaration
int main()
{
int m = 25;
int n = 100;
printf("m is %d, n is %d\n", m, n);
swap(&m, &n); //calling a function
printf("m is %d, n is %d\n", m, n);
return 0;
}
void swap (int *a, int *b) //function implementation
{
int temp;
temp = *a;
*a = *b;
*b = temp;
}
in the above program comment the line void swap (int *a, int *b); then program run file but give one suggestion here facing the issue why here give the suggestion
main.c:10:3: warning: implicit declaration of function ‘swap’ [-Wimplicit-function-declaration]
main.c:16:6: warning: conflicting types for ‘swap’
main.c:10:3: note: previous implicit declaration of ‘swap’ was here
m is 25, n is 100
m is 100, n is 25
when I m uncomment this line void swap (int *a, int *b); then program run fine
#include<stdio.h>
void swap (int *a, int *b); //function declaration
int main()
{
int m = 25;
int n = 100;
printf("m is %d, n is %d\n", m, n);
swap(&m, &n); //calling a function
printf("m is %d, n is %d\n", m, n);
return 0;
}
void swap (int *a, int *b) //function implementation
{
int temp;
temp = *a;
*a = *b;
*b = temp;
}
problem is why not allowed this one in the above program
void swap (int, int); //function declaration
and give the error
error:
conflicting types for ‘swap’
void swap (int *a, int *b) //function implementation
but in book example allowed
void swap (int *a, int *b);
This here before main() is called function signature that should match function declaration
below.
void swap (int *a, int *b){
...
}
But the same function signature can be written as
void swap (int *, int *);
And not
void swap (int , int ); //This will result in type mismatch
I guess this is what you are asking.
NOTE: Forward declaration is important here which lets you declare first and define later helps maintain code readability.
These are two functions I used for swapping variables using pointers while performing selection sort using the function selectionSort(int *,int). But after sorting, some elements of the array become zero.
void selectionSort(int *x,int len){
int i,j,max;
for(i=len-1;i>=0;i--){
max = 0;
for(j=1;j<=i;j++){
if(x[j]>x[max]){
max = j;
}
}
swap(x+max,x+i);
}
}
void swap(int *a,int *b){
//This one works perfectly
int temp;
temp=*b;
*b=*a;
*a=temp;
}
void swap(int *a,int *b){
//This one gives unexpected results
*a=*a+*b;
*b=*a-*b;
*a=*a-*b;
}
Swapping two integers using arithmetic operators can result in integer overflow. Better to stick with the traditional approach.
BTW, you can use old school bitwise XOR operator for swapping (used to swap values in registers) but it would not give you any benefit over the method that uses a temporary variable. These days compiler are smart enough to optimize the code.
if (*a == *b) // If both integers are same then do not perform swap operation
return;
*a ^= *b;
*b ^= *a;
*a ^= *b;
The second swap algorithm does not work if both pointers point to the same variable which happens here if max == i:
swap(x+max,x+i);
Demonstration:
#include <stdio.h>
#include <string.h>
#include <assert.h>
void swap(int *a, int *b) {
//This one works perfectly
int temp;
temp = *b;
*b = *a;
*a = temp;
}
void swap_KO(int *a, int *b) {
//This one gives unexpected results
*a = *a + *b;
*b = *a - *b;
*a = *a - *b;
}
int main()
{
int a = 22, b = 33;
swap(&a, &b);
assert(a == 33 && b == 22); // OK
swap_KO(&a, &b);
assert(a == 22 && b == 33); // OK
swap(&a, &a); // OK
assert(a == 22);
swap_KO(&a, &a); // won't work
assert(a == 22);
}
I'm trying to get my head around pointers and I'm trying to write a program which will swap two numbers using pointers. However, I'm getting the error as stated in the title. Here is my code:
//This program swaps two numbers using pointers
#include <stdio.h>
void swap(*val1, *val2);
int main() {
int num1, num2;
int *pNum1, *pNum2;
printf("Enter number 1:\n");
scanf("%d", &num1);
printf("Enter number 2:\n");
scanf("%d", &num2);
pNum1 = &num1;
pNum2 = &num2;
printf("Numbers not swapped: %d, %d\n", *pNum1, *pNum2);
swap(pNum1, pNum2);
return 0;
}
void swap(*val1, *val2) {
int temp;
temp = val1;
val1 = val2;
val2 = temp;
printf("Numbers swapped: %d, %d\n", *val1, *val2);
return;
}
void swap(*val1, *val2);
should be
void swap(int *val1, int *val2);
You should then pass
swap(&num1,&num2);
If you pass a pointer then you are passing a copy of it.You need to pass the address. No need of having pointers in the calling function you can directly pass the address of the variables.
void swap(int *p,int *q)
{
int t = *p;
*p = *q;
*q = t;
}
This code is working but its not exactly what I want. Is anyone have any idea how to make it correct and without q sort?. The idea is to understand how to use pointers.
The three numbers should be random between -3 and 12. The code below is something similar, and the closest I have found. Any help would be much appreciated. Thanks in advance!!.
#include <stdio.h>
#include <stdlib.h>
//functions
int compare(const void *a, const void *b)
{
const int *ia = a;
const int *ib = b;
if (*ia < *ib)
return -1;
else if (*ia > *ib)
return +1;
return 0;
}
//qsort function
void sort3(int *a, int *b, int *c)
{
int temp[3];
temp[0] = *a;
temp[1] = *b;
temp[2] = *c;
qsort(temp, 3, sizeof(int), &compare);
*a = temp[0];
*b = temp[1];
*c = temp[2];
}
//random function
int rand_int(int a, int b)
{
return rand()%(b-a+1)+a;
}
int main(void)
{
//declaration of variables
int a,b,c;
int rand_int(int a, int b);
srand(time(0));
a = rand_int(-3,12);
b = rand_int(-3,12);
c = rand_int(-3,12);
printf("%i %i %i\n", a, b, c);
sort3(&a, &b, &c);
printf("%i %i %i\n", a, b, c);
return 0;
}
You don't need the compare() function if you don't want to use qsort().
You can rewrite sort3() like this:
void compare_and_swap(int *a, int *b) {
int t;
if (*a > *b) {
t = *a;
*a = *b;
*b = t;
}
}
void sort3(int *a, int *b, int *c) {
compare_and_swap(a, b);
compare_and_swap(a, c);
compare_and_swap(b, c);
}
This is actually a "bubble sort".
This is a lot of trouble to go for to sort 3 integers. Use if statements.
If the goal is actually to understand pointers, they seem intimidating but they're not so bad. Basically, they're a number that happens to be an address. You can manipulate like them numbers, but if you dereference them (with *), you can get the value there. This cuts both ways, though, because there's not much stopping you from dereferencing a value - which probably crashes your program (or more scarily, maybe not).
As long as you keep in mind what's an address and what's a value, you should be OK.