as you said that the function will receive the address of the array defined in main function so isnt it going to use the same memory space or in simply the same array?
I have a huge doubt now! Firstly u have written int *p =a;(u said it stores the adress of a), but nextly u said *p gives the values of the adress...im confused
int *p defines p as a pointer, here the (int *) says that p is a special variable that stores address of some int variable. But when you do printf("%d", *p) --> here * means value at address stored by p. Hope u got the difference.
Excellent, explained precisely, good visual demos
Excellent!!!! It really help me a lot.
Sir am very grateful that you helped me understand C programming.
Congratulations from Brazil.
Gracias !
Master class
Perfect...
Real talent
Awesome explanation. Thank you. Love from Bangladesh.
Thank you, Sir, excellent way of teaching.....
very well explained!! thank you, sir.
wow!
excellent explaination sir.
if possible just give a number to your videos. It will be helpful to see in order. and thank you
I got what i was looking for .thank you.
Very well made!
You're actually the best
your tutorials are good.if possible please provide tutorials on jsp&servlet.
Kanishka Das it's very understandable for the way your giving the examples thank u😁😀
In 5:22, address of a is 1000, so address of b is 1000 (but 2000); pass reference.
very well explained sir
as you said that the function will receive the address of the array defined in main function so isnt it going to use the same memory space or in simply the same array?
as the array name which is defined in the main function will pass as an argument to the display function?
#include
#include
using namespace std;
int pause = 0;
void display(int b[5], int size){
for (int i = 0; i < size; i++){
printf("
%d",b[i]," at:");
printf(" %d", &b[i]);
}
}
void main(){
int num[] = { 2, 4, 6, 8, 10 };
printf("
In main function.");
for (int i = 0; i < 5; i++){
printf("
%d", num[i], " at:");
printf(" %d", &num[i]);
}
display(num,5);
scanf("%d", pause);
}
hello sir I am using turbo c++ editor in window 10 and int is incremented by 2 byte and you increment by 4 why?
Better explaination
Respected Sir , please upload on Pointer to 2d Array... Please sir
I have a huge doubt now! Firstly u have written int *p =a;(u said it stores the adress of a), but nextly u said *p gives the values of the adress...im confused
int *p defines p as a pointer, here the (int *) says that p is a special variable that stores address of some int variable. But when you do printf("%d", *p) --> here * means value at address stored by p. Hope u got the difference.
Thank you @bbarters
Sir pls post the let us c book solutions pls with in 2days pls sir plsssss
You are ✋+