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);string length c
#include <stdio.h>
#include <string.h>
int main()
{
char a[20]="Program";
char b[20]={'P','r','o','g','r','a','m','\0'};
// using the %zu format specifier to print size_t
printf("Length of string a = %zu \n",strlen(a));
printf("Length of string b = %zu \n",strlen(b));
return 0;
}array length c
int prices[5] = { 1, 2, 3, 4, 5 };
int size = sizeof prices / sizeof prices[0];
printf("%u", size); /* 5 */c how to find size of array
int a[17];
size_t n = sizeof(a)/sizeof(a[0]);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 7Copyright © 2021 Codeinu
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