I'm with this doubt: how to get the size of a char array in this case:
#include<stdio.h>
void f(char * x)
{
printf("Size %d\n", sizeof(x)/sizeof(char));
}
main()
{
char x[5] = {'a', 'e', 'i', 'o', 'u'};
f(&x[0]);
}
Contrary to my expectations, I'm receiving 8 rather than 5 or even 6. What is wrong here?
Thanks!
sizeof(x) in your code will return the size of pointer char *x and not the size of the char array that x is pointing on
and the size of pointer in your 64-bits system is 8. and for 32-bits system the size of pointer is 4
Here, sizeof() is returning the size of the pointer, not the size of the original array.
The only way for f() to know the size of the array pointed to by the char* is for it to be told by the caller:
void f(char * x, size_t size)
{
...
}
main()
{
char x[5] = {'a', 'e', 'i', 'o', 'u'};
f(x, sizeof(x) / sizeof(x[0]));
}
You can not. sizeof returns the size of a pointer.
Store the size of your array in a variable and pass it too.
I prefer this:
void f(char *x, int size)
{
// ...
}
main()
{
char x[5] = {'a', 'e', 'i', 'o', 'u'};
f(x, 5);
}
sizeof(x) gives you size of the char pointer not the size of the array. char * is a pointer to a char. If you dochar a = 'A'; f(&a); it is still valid. char * is not designed to point to only char arrays, so sizeof(x) returns size of the pointer and not what it is pointing at.
You get it like this
void f(char * x, int size_of_array)
{
printf("Size %d\n", size_of_array);
}
main()
{
char x[5] = {'a', 'e', 'i', 'o', 'u'};
f(&x[0], 5);
}
Once you pass an array it decays into a pointer of that type, and you loose the ability to get the size of the array via the sizeof macro. You need to pass the number of elements. If your array is of numeric type you can always pass the size of an array as the first element:
void f(char * x)
{
printf("Size %d\n", x[0]);
}
main()
{
char x[6] = {6, 'a', 'e', 'i', 'o', 'u'};
f(&x[0]);
}
But of course in this case there's extra overhead to updating that element to make sure it matches what you expect.
%dcorresponds to anintargument. Thesizeofoperator produces asize_t, not anint. The argument mismatch produces undefined behaviour. Use%zu, instead. 2.sizeof (char)is always one. Dividingsizeof (x)by one producessizeof (x). Clean up your code by removing this unnecessary division-by-one operation.