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);get length of array in c
int a[17];
size_t n = sizeof(a)/sizeof(a[0]);size of array
#include <stdio.h>
#include <stdlib.h>
void printSizeOf(int intArray[]);
void printLength(int intArray[]);
int main(int argc, char* argv[])
{
int array[] = { 0, 1, 2, 3, 4, 5, 6 };
printf("sizeof of array: %d\n", (int) sizeof(array));
printSizeOf(array);
printf("Length of array: %d\n", (int)( sizeof(array) / sizeof(array[0]) ));
printLength(array);
}
void printSizeOf(int intArray[])
{
printf("sizeof of parameter: %d\n", (int) sizeof(intArray));
}
void printLength(int intArray[])
{
printf("Length of parameter: %d\n", (int)( sizeof(intArray) / sizeof(intArray[0]) ));
}size of array
size_t n = sizeof myArray / sizeof *myArray;Copyright © 2021 Codeinu
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