size of an array c
int a[20];
int length;
length = sizeof(a) / sizeof(int);size of an array c
int a[20];
int length;
length = sizeof(a) / sizeof(int);array length c
int prices[5] = { 1, 2, 3, 4, 5 };
int size = sizeof prices / sizeof prices[0];
printf("%u", size); /* 5 */get length of array in c
int a[17];
size_t n = sizeof(a)/sizeof(a[0]);size of an array c
int a[]= { 1, 2, 3, 4, 5, 6, 7 };
int length = sizeof(a) / sizeof(a[0]); //return 7passing the length of an int array c
//I suggest using strlen (of #include<string.h>)
//if you want only the size of the occupied index
int prices[5] = { 1, 2, 3, 4, 5 };
int size = strlen((char*)prices); // convert it to string
printf("%d", size+1); // Just add 1 that's itsize of an array in c
size_t n = sizeof myArray / sizeof *myArray;Copyright © 2021 Codeinu
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