Array of Pointers | Logical Programming in C | Naresh IT

Array of Pointers | Logical Programming in C | Naresh IT

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[sd]

At 01:30 'a ' is a variable storing the base address of first element of array but what is the address of variable 'a'???🙏

[eren_n]

4:56 but we can't write array a[i] as a+i bcoz it's not a pointer right?

[gnaneshnayakbadavath]

very good explanation sir, but write with blue marker, with black marker we not visible clarityly

[ammamanusha]

Ur explanation is excellent sir...!👌Thnqqq🙏 But am confusing while am practicing

[unnamlalitha]

Thank u sir for ur explanation it is very easy to understand with ur explanation

[nishantprabhakar]

I am not able to understand this *(p+i)=a+i
As in the second for loop 2046+1 will it not become 2047. I am getting confused here since we avoiding index.

[bunnynikhil]

Thank u sir .... explained superly sir

[Sar]

Sir, your explanation is awesome but try to write little bit dark over the board and it is been tough to catch the writings. Hope u may not mistaken me...

[nindo]

sir why we are not using & operator in front of a+1 for storing the reference

[prabhuchaitanya]

we need to print *p[i] thanks for explanation sir

[kavyaroyal]

Sir u told in this video, can we store different base address of arrays
How can assign to pointer arrays

[iplwinassets]

If the address of a[0] is 4046, , then the address of a[1] should be 4050, , not 4048, , because it is an integer type, , the incremen should be 4 bytes

[food]

sir pen accha use karo clear nhi dekh raha h

[AtulSingh-dovu]

Sir teach array of pointers using dynamic memory allocation.. Please

[Odo_ela_sabil_rabek]

i think it should be like this
*(p+i)=(int*)(a+i)
because the name of the array 'a' if i write a+1 that's mean it increases by array size i.e(sizeof(int)*5) am i right?

[hridyanshpareek]

TO PRINT ELEMENTS STORED IN ARRAY OF POINTERS.

#include <stdio.h>

int main()
{

int a[5] = {1, 2, 3, 4, 5};

int *p[5];
for(int i = 0; i<5; i++)
{
*(p+i) = a+i;//p[i] = &a[i];
}

for(int i = 0; i<5; i++)
{
printf("%d\n", *(*(p+i)));
}

return 0;
}