dereference increment

[junior notch]

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@teal raptor

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[teal raptor]

char a[2] = "0";
*a = '1';```

this sets a[0] to 1 ... how would i set the next index in this way

[atomic steeple]

a[i] = x;

btw " is string use ' for char

[teal raptor]

*a goes to the first index of the array

[atomic steeple]

yeah

[teal raptor]

can I set the next index by incrementing that somehow?

instead of the normal way

[atomic steeple]

yeah

you can increment it with sizeof(char)

[atomic steeple]

but if you have int arr[3]; you have to increment with sizeof(int)

[teal raptor]

I see

*a + sizeof(char) = '4';

error: lvalue required as left operand of assignment

[runic cape]

dont delete questions

they may be interesting for other people

[teal raptor]

yeah I Read the automated message 🙂

not sure who marked this as solved

[atomic steeple]

;compile

#include <cstdio>
int main(void) {
    int arr[10] = {0, 69};
    printf("%d %d\n", arr[1], *(arr + 1));
    *(arr + 1) = 50;
    printf("%d %d\n", arr[1], *(arr + 1));
}

[modest geode]

Program Output

69 69
50 50

MACAYCZ#6241 | 41ms | c++ | x86-64 gcc 12.2 | godbolt.org

[atomic steeple]

you dont have to write sizeof(char) because compiler do the thing for you, also you have to increment the pointer before

[teal raptor]

compiler does what thing?

*(a + sizeof(char)) = '4';

works

[atomic steeple]

it increments the address with sizeof the elements

[atomic steeple]

yeah but dont write the sizeof

[teal raptor]

oh so you can just do + 1

ty

!solved

[junior notch]

Message is already solved

[atomic steeple]

;compile

#include <cstdio>
int main(void) {
    int arr[10];
    printf("%p %p\n", arr+0, arr+1);
}

[modest geode]

Program Output

0x7fff64ecde70 0x7fff64ecde74

MACAYCZ#6241 | 46ms | c++ | x86-64 gcc 12.2 | godbolt.org

[atomic steeple]

the compiler offset the address by sizeof(int)

but you don't have to

[teal raptor]

ok

got it

ty

[atomic steeple]

this works the same way but the compiler offsets the pointer only by 1

int arr[10] = {0, 69};
printf("%d\n", *((char*)arr + sizeof(*arr)));

[teal raptor]

[atomic steeple]

because i typecast it to char array and char have size 1 byte

;compile

#include <cstdio>

int main() {
  int arr[10] = {0, 69};
  printf("%d\n", *((char*)arr + sizeof(*arr)));
  return 0;
}

[modest geode]

Program Output

69

MACAYCZ#6241 | 54ms | c++ | x86-64 gcc 12.2 | godbolt.org

[atomic steeple]

typecast dont cost any computational time