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cryptod · Sep 2nd, 2007, 9:42 PM

Its impossible to do it without passing the length of thr array as a parameter.
here is a possible solution :

void foo(unsigned *p, int length)
{
    cout << length << '\n';
}

int main (int argc, const char * argv[]) {
    unsigned p[] = {1,2,3,4,5,6,7,8,9,0};
int length=sizeof(p)/sizeof(p[0]) ;
    foo(p,length);
    return 0;
}

Good luck

Sep 2nd, 2007, 9:42 PM	#7
cryptod Newbie Join Date: Sep 2007 Location: Algeria Posts: 2 Rep Power: 0	Its impossible to do it without passing the length of thr array as a parameter. here is a possible solution : (Toggle Plain Text) void foo(unsigned p, int length) { cout << length << '\n'; } int main (int argc, const char argv[]) { unsigned p[] = {1,2,3,4,5,6,7,8,9,0}; int length=sizeof(p)/sizeof(p[0]) ; foo(p,length); return 0; } void foo(unsigned p, int length) { cout << length << '\n'; } int main (int argc, const char argv[]) { unsigned p[] = {1,2,3,4,5,6,7,8,9,0}; int length=sizeof(p)/sizeof(p[0]) ; foo(p,length); return 0; } Good luck