const char *array[] = {"ax","bo","cf"};
tried
printf("size of array = %lu\n", sizeof(const char*));
result != 3
also
printf("size of array = %lu\n", sizeof(array));
result != **DESIRED ANSWER** = 4
NOTE... I have read related questions on here but none had a relation with my question......
To get the size of a const char pointer:`
printf("%zu\n", sizeof(const char *));
To get the size of the array array[]:
const char *array[] = {"ax","bo","cf"};
printf("%zu\n", sizeof array);
To get the number of elements in the array array[], divide the size of the array by the size of an array element.
const char *array[] = {"ax","bo","cf"};
// Size of array/size of array element
printf("%zu\n", sizeof array / sizeof array[0]);
// expect 3
sizeof array / sizeof array[0] or sizeof array / sizeof *array. BTW I think the third example has typo. Should be array instead of a.
sizeof array / sizeof array[0] or sizeof array / sizeof *array is easy. Every group should employ a coding style that settles style issues like this and then use the agreed upon one. I prefer [] at it emphasizes the array nature of the object.sizeof array/sizeof(const char*) style as that is harder to code, review and maintain. To know the type is correct requires cross checking with the declaration. This is especial a nuisance when the type is defined in a field of some structure located in a header file.anyType array[3] = {};
sizeof(array[0]) //will return the size of one element of any array
sizeof(array) //will return the storage requirement of the entire array
(sizeof(array)/sizeof(array[0])) //will return the number of elements in the array
But remember that sizeof() resolves at compile time so it only works if you have the actual array variable, not a pointer to an element. Meaning you can't pass the array into a function and then take the sizeof() of the passed parameter unless you declare the parameter to be an array of fixed size.
sizeof also applies to variable length arrays, a run-time determination.
you need to first find out the size of const char * then you need to divided by this to size of you array like this :
#include <stdio.h>
#include<string.h>
int main()
{
const char *array[] = {"ax","bo","cf"};
printf("%d",sizeof(array)/sizeof(const char*));
return 0;
}
const,volatile) is the same as the size of a non-qualified pointer.constqualifier can make a pointer point to text (eprom), so using a different access than normal ram and sometimes a different size of the pointer.const,volatile, andrestrictqualifiers. The qualified or unqualified versions of a type are distinct types that belong to the same type category and have the same representation and alignment requirements.