Thank you! your explanations are quite vivid and clear. Kindly try to mention at least the main headings in English. Most of the program headings are in English but some weren't. However, overall a great place to learn coz I love your mannerisms while explaining and making complex programs simple to apprehend.
Sir onek video te like cmnt korte mon thake na ,video dekhi r shate shate code kore feli tay amon ta hoi ...apnar dara onek kisu Allah shikhaitase duah kori apni valo thaken r valo valo tutorial amader k den
/* C program to swap two number using call by reference */ #include void swapping(int *x, int *y); int main() { int x, y; printf("Enter two numbers: "); scanf("%d %d", &x, &y); printf("before swapping of x = %d ", x); printf("before swapping of y = %d ", y); swapping(&x, &y); return 0; } void swapping(int *x, int *y) { int *p1,*p2; p1=&x; p2=&y; int temp; temp = *p1; *p1= *p2; *p2= temp; printf("after swapping of x = %d ", *x); printf("after swapping of y = %d ", *y); }
can i have a ques answer we know the * sign define that it mean in a pointer it will store the value of the variable but in user defined function we returned the address of the variable how it is printing the value
cause pointers work with memory address so when you swapped the value of &x(memory address of x ) and &y it changed the value of those memory addresses so no matter from where you change it and where you print it from, the value for the specific memory will be same.
Thank you! your explanations are quite vivid and clear. Kindly try to mention at least the main headings in English. Most of the program headings are in English but some weren't. However, overall a great place to learn coz I love your mannerisms while explaining and making complex programs simple to apprehend.
Sir onek video te like cmnt korte mon thake na ,video dekhi r shate shate code kore feli tay amon ta hoi ...apnar dara onek kisu Allah shikhaitase duah kori apni valo thaken r valo valo tutorial amader k den
This tutorial is very nice and beautiful..
vaiya apni je fuction ta use koresen seita to void silu tahole value gulo main funtion e aslo ki kore
excellent tutorial...... thanks brother
/*
C program to swap two number using call by reference */
#include
void swapping(int *x, int *y);
int main()
{
int x, y;
printf("Enter two numbers: ");
scanf("%d %d", &x, &y);
printf("before swapping of x = %d
", x);
printf("before swapping of y = %d
", y);
swapping(&x, &y);
return 0;
}
void swapping(int *x, int *y)
{
int *p1,*p2;
p1=&x;
p2=&y;
int temp;
temp = *p1;
*p1= *p2;
*p2= temp;
printf("after swapping of x = %d
", *x);
printf("after swapping of y = %d
", *y);
}
Thank you this video is really helpful
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can i have a ques answer
we know the * sign define that it mean in a pointer it will store the value of the variable
but in user defined function we returned the address of the variable
how it is printing the value
vaia malloc, calloc and realloc and free niye tutorial dile upokrito hotam....
Thanks vai❤
Great Explain ❤️
Sir, plz Pass by value, pass by reference ta dekhaiyen. Using pointer
vaiya apni je fuction ta use koresen seita to void silo taile value gula main function e aslo kmne?
cause pointers work with memory address so when you swapped the value of &x(memory address of x ) and &y it changed the value of those memory addresses so no matter from where you change it and where you print it from, the value for the specific memory will be same.
thanks a lot
Thanks
Thank you
Thank you via
function tah main er pore declare krle hbe na?
Thank You sir
tnq
User input thakle shey khetre ki korbo?
"permutation of a string using pointer " please give me a solution of this problem..
Ami run korle asena folafol
Thank you