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int main()
{
int i = 10;
int *p = &i;
foo(&p);
printf ("%d ", *p);
printf ("%d ", *p);
}
void foo(int **const p)
{
int j = 11;
*p = &j;
//Printing the vlue
printf("%d ", **p);
}
When foo returns, the pointer p in main points to a local variable that existed during the execution of foo. Since foo has ended, de-referencing that pointer invokes undefined behaviour. Therefore your program can output, or indeed do, anything.
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This is the code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int appar(char c[], char x);
int main() {
char c[] = "hello everyone!";
int b = appar(c, 'h');
printf("nbr of h is %d ", b);
return 0;
}
int appar(char c[], char x) {
int i = 0, cmpt = 0;
int q = strlen(c);
for (i; i < q; i++) {
if (c[i] == 'x')
cmpt++;
}
return cmpt;
}
I run and compile the program, but I receive "nbr of h is 0".
What's the wrong in this code?
Change c[i]=='x' to c[i]==x
You want to compare with the variable x, not the character constant 'x'
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I am trying to print the sum and diffrence of numbers using pointers but i am getting only sum as output.
#include <stdio.h>
#include<math.h>
void update(int *a,int *b)
{
int sum,sub;
sum = *a + *b;
printf("",sum);
sub = abs(*a - *b);
printf("",sub);
}
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;
}
If I have understood correctly (taking into account the function name) you mean something like the following
#include <stdio.h>
#include <stdlib.h>
void update( int *a, int *b )
{
*a += *b;
*b = abs( *a - *b - *b );
}
int main(void)
{
int a, b;
scanf( "%d%d", &a, &b );
update( &a, &b );
printf( "a = %d, b = %d\n", a, b );
return 0;
}
If for example to enter two values 20 and 10 then the output will look like
a = 30, b = 10
Pay attention to that these calls of printf in your function
printf("",sum);
printf("",sub);
do not make sense.
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I'm trying to write a program in which there are two pointers that point to the min and the max value of a 100-values array. Why does it give me error?
#include <stdio.h>
int main(void){
int array[100], i;
for(i=0; i<100; i++)
array[i]=i;
int *ptr1, *ptr2, flag=0;
ptr1 = &array[0];
ptr2 = &array[0];
while(!flag){
for(i=0;i<100;i++){
if(*ptr1 > array[i]){
ptr1 = &array(i);
break;
}else if(*ptr2 < array[i]){
ptr2 = &array(i);
break;
}
}
if(i==100)
flag=1;
}
printf("%d %d", *ptr1, *ptr2);
}
main.c:13:25: error: called object ‘array’ is not a function or function pointer
The statement
ptr1 = &array(i);
should be corrected to
ptr1 = &array[i];
So does the
ptr2 = &array(i);
This question already has answers here:
returning a local variable from function in C [duplicate]
(4 answers)
Closed 7 years ago.
Why is the First Printing Statement in main(), printing 11 ?
#include<stdio.h>
void foo(int ** p){
int j = 11;
*p = &j;
printf("%d ", **p); //Printing 11
}
int main(){
int i = 10;
int *p = &i;
foo(&p);
printf("%d ", *p); //Printing 11
printf("%d ", *p); //Printing Random value
return 0;
}
Inside foo() , you're assigning the address of a automatic local variable j to *p. After foo() has finished execution, j does not exist anymore and thus, using (derererencing) p furthermore in main() invokes undefined behavior.
Now, the output of UB is, well, undefined.
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Closed 5 years ago.
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Can you help me by telling me what is wrong with this? Why is the swap function not working?
void swap(int a[], int b, int c) {
int temp = a[b];
a[b] = a[c];
a[b] = temp;
}
void bubble1 (int a[], int N){
int i;
for(i=0;i<N-1;i++){
if(a[i]>a[i+1]){
swap(a,i,i+1);
}
}
}
void main() {
int N = 11;
int a[12]={5,3,12,4,25,10,14,35,2,8,13};
bubble1 (a,N);
int i;
for(i = 0; i < N; i++){
printf("%d\n",a[i]);
}
}
If I don't use the swap function and do the swapping manually in the "bubble" function it works. However if I use the swap it doesn't work, even though it's exactly the same. What am I doing wrong here?
int temp = a[b];
a[b] = a[c];
a[b] = temp;
Simple typo, you are assigning to a[b] twice. The second one should be a[c]