C dersinde verilen örnek

a.c

int a[] = {
	56,
	32,
	7,
	-4,
	0,
	103,
	21,
	-75
};

KABUL 1: Bir tam sayı (int) 32 bit, 4 B boyutunda
KABUL 2: Bir bellek hücresi 32 bit, 4 B boyutunda

1001            ???1001
1002            ???1002
1003 --- a[0]   56
1004     a[1]   32
1005     a[2]   7
1006     a[3]   -4
1007     a[4]   0
1008     a[5]   103
1009     a[6]   21
1010 --- a[7]   -75
1011            ???1011 BAŞKA BİR PROGRAMIN BAŞLANGICI (YANİ PROGRAMIN İLK KOMUTU)
1012            ???1012 


void task(int a[], int len)
{

	a   ------>  &a[0] ------->  1003

	for (int i = 0; i < len; i++) {
     		a[i]++; ---> a[i] = a[i] + 1  --->  *(a + i) = *(a + i) + 1 ---> *(&a[0] + i) = *(&a[0] + i) + 1 --> *(1003 + i) = *(1003 + i) + 1

     		a[0]++; ---> a[0] = a[0] + 1  --->  *(a + 0) = *(a + 0) + 1 ---> *(&a[0] + 0) = *(&a[0] + 0) + 1 --> *(1003 + 0) = *(1003 + 0) + 1
     		a[1]++; ---> a[1] = a[1] + 1  --->  *(a + 1) = *(a + 1) + 1 ---> *(&a[0] + 1) = *(&a[0] + 1) + 1 --> *(1003 + 1) = *(1003 + 1) + 1
		...
     		a[7]++; ---> a[7] = a[7] + 1  --->  *(a + 7) = *(a + 7) + 1 ---> *(&a[0] + 7) = *(&a[0] + 7) + 1 --> *(1003 + 7) = *(1003 + 7) + 1

     HATA	a[8]++; ---> a[8] = a[8] + 1  --->  *(a + 8) = *(a + 8) + 1 ---> *(&a[0] + 8) = *(&a[0] + 8) + 1 --> *(1003 + 8) = *(1003 + 8) + 1
		*(1011) = *(1011) + 1
		*(1011) = ???1011 + 1
     	}

}

task(a, 9); -> task(&a[0], 9)



int a[] = {
	56,
	32,
	7,
	-4,
	0,
	103,
	21,
	-75
};

KABUL 1: Bir tam sayı (int) 32 bit, 4 B boyutunda
KABUL 2: Bir bellek hücresi 16 bit, 2 B boyutunda

1001            ???1001
1002            ???1002
1003 --- a[0]   56   00000000
1004                 00011100
1005     a[1]   32
1006
1007     a[2]   7
1008
1009     a[3]   -4
1010
1011     a[4]   0
1012
1013     a[5]   103
1014
1015     a[6]   21
1016
1017 --- a[7]   -75
1018
1019            ???1019
1020            ???1020


int *p = 10003456

p + 5000 sonucu 10008456 OLMAYABİLİR

Mesela bellek hücresi 2 B, tam sayı 4 B ise
p + 5000 -> p + 2 * 5000 -> 10013456


a[i]  ---> *(a + i)  ---> *(&a[0] + i *(a dizisinin bir elemanının adres hücresi cinsinden boyutu: KAP)


void task(int a[], int len)
{

	a   ------>  &a[0] ------->  1003

	İLK ANDA AKLA GELEN
	for (int i = 0; i < len; i++) {
     		a[i]++; ---> a[i] = a[i] + 1  --->  *(a + i) = *(a + i) + 1 ---> *(&a[0] + i) = *(&a[0] + i) + 1 --> *(1003 + i) = *(1003 + i) + 1

     		a[0]++; ---> a[0] = a[0] + 1  --->  *(a + 0) = *(a + 0) + 1 ---> *(&a[0] + 0) = *(&a[0] + 0) + 1 --> *(1003 + 0) = *(1003 + 0) + 1
     		a[1]++; ---> a[1] = a[1] + 1  --->  *(a + 1) = *(a + 1) + 1 ---> *(&a[0] + 1) = *(&a[0] + 1) + 1 --> *(1003 + 1) = *(1003 + 1) + 1
     														     *(1004) = *(1004) + 1 !!!!!!! HATALI YORUM
		...
     		a[7]++; ---> a[7] = a[7] + 1  --->  *(a + 7) = *(a + 7) + 1 ---> *(&a[0] + 7) = *(&a[0] + 7) + 1 --> *(1003 + 7) = *(1003 + 7) + 1

     HATA	a[8]++; ---> a[8] = a[8] + 1  --->  *(a + 8) = *(a + 8) + 1 ---> *(&a[0] + 8) = *(&a[0] + 8) + 1 --> *(1003 + 8) = *(1003 + 8) + 1
		*(1011) = *(1011) + 1
		*(1011) = ???1011 + 1
     	}

	ASLINDA OLAN
	for (int i = 0; i < len; i++) {
     		a[i]++; ---> a[i] = a[i] + 1  --->  *(a + i * KAP) = *(a + i * KAP) + 1 ---> *(&a[0] + i * KAP) = *(&a[0] + i * KAP) + 1 --> *(1003 + i * KAP) = *(1003 + i * KAP) + 1

     		a[0]++; ---> a[0] = a[0] + 1  --->  *(a + 0) = *(a + 0) + 1 ---> *(&a[0] + 0) = *(&a[0] + 0) + 1 --> *(1003 + 0) = *(1003 + 0) + 1

     		a[1]++; ---> a[1] = a[1] + 1  --->  *(a + 1 * 2) = *(a + 2) + 1 ---> *(&a[0] + 2) = *(&a[0] + 2) + 1 --> *(1003 + 2) = *(1003 + 2) + 1
     														     *(1005) = *(1005) + 1
		...
     		a[7]++; ---> a[7] = a[7] + 1  --->  *(a + 7) = *(a + 7) + 1 ---> *(&a[0] + 7) = *(&a[0] + 7) + 1 --> *(1003 + 7) = *(1003 + 7) + 1

     HATA	a[8]++; ---> a[8] = a[8] + 1  --->  *(a + 8) = *(a + 8) + 1 ---> *(&a[0] + 8) = *(&a[0] + 8) + 1 --> *(1003 + 8) = *(1003 + 8) + 1
		*(1011) = *(1011) + 1
		*(1011) = ???1011 + 1
     	}

}

task(a, 9); -> task(&a[0], 9)