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Why compiler doesn't pass size of array char *arr[] in parameters? I wanted to get get size of array passed by parameter but I guess it doesn't work because even char *a[] is char ** my question is why is it and can I make it work?
#include <stdio.h>
#include <stddef.h>
#include <stdio.h>
template<class T, size_t len>
constexpr size_t lengthof(T(&)[len])
{
return len;
}
void printarr(const char *a[]);
int main()
{
const char *a[] = { "aba", "bd", "cd" };
printarr(a);
}
void printarr(const char *a[])
{
for(size_t i = 0, c = lengthof(a); i < c; i++) {
printf("str = %s\n", a[i]);
}
}
719
The reason you can't pass an array by value is because there is no specific way to track an array's size such that the function invocation logic would know how much memory to allocate and what to copy. You can pass a class instance because classes have constructors. Arrays do not.
If you create an array by initializing its values directly, the size will be the number of elements in it. Thus the size of the array is determined at the time of its creation or, initialization once it is done you cannot change the size of the array.
Because `sizeof(a)` is referring to `a` as a pointer rather than an array. In C an array's length must be passed separately because it is not remembered within the array. in short.. when you pass a[] in a function it will decay into pointer to an int .
Passing the array size tells the function where the bounds are so you can choose not to go beyond them.
You can make it work by using the same trick you used in your lengthof function template.
template<size_t len>
void printarr(const char* (&a)[len])
{
for(size_t i = 0, c = lengthof(a); i < c; i++) {
printf("str = %s\n", a[i]);
}
}
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