Method Invoking

[coisox 0]

Using c++:

int main()

{

cout << " Sum is: " << funct(5);

}

int funct(int y)

{

if (y==1 )

return 1;

else

return y++;

}

Dari program nih, secara logiknye, bagi aku jawapan ialah "sum is: 6". Tapi yg buat aku tak paham, bila aku compile jawapan ialah "sum is: 5". Boley sesaper tolong explain kat aku, kenapa proses penambahan tu tak berlaku. Tak pahamlah!

Proses penambahan akan berlaku sekiranya the last two statement (program kat atas) tukar kepada:

else

y++;

return y

Tolooooooonggggg.......

[encik pot pet 0]

cuba guna,

return ++y;

[Paralys3r 0]

return y++;

basically means

return y;
y++; //or ++y; it doesn't matter now.

[mchammer 0]

Ini dinamakan pre/post increment.

Contoh:

1) jumlah = y++;

Assign value y ke variable jumlah, lepas tu baru lakukan proses increment.

2) jumlah = ++y;

Lakukan proses increment, lepas tu simpan hasil increment y ke variable jumlah

pre/post decrement pun macam tu jugak.

So. Kalau buat:

return y++;

Maknanya kita just return value y sebelum proses increment.

[encik pot pet 0]

return y++;

basically means

return y;
y++; //or ++y; it doesn't matter now.

it does matter. consider

funct(int y)
{
    int i = 2;
    int j = 3;
    return y+i+j; // if y = 5, return value should be 10 
}

compiler akan accumulate hasil tambah semua tu baru return,

sebab y++ tak incremented sebab operator tu post-increment. kalau pre-increment, program akan execute increment value dulu baru return call with value in eax (x86).

[Paralys3r 0]

please use the context above.

after return statement.. it really doesn't matter whether you use post- or pre- increment. Maknanya

return y; 
y++;

dan

return y;
++y;

is just the same thing. Kalau convert directly ke machine code ia akan jadi (secara ringkasnya )

mov eax, [esp+4]
cmp eax, 1
jnz __else
ret
__else:
ret
inc eax

Bermakna inc eax TAK AKAN DILAKSANAKAN. Anything after return statement is useless kalau takde branch pergi ke sana untuk dilaksanakan.

Edited September 8, 2007 by Paralys3r

[encik pot pet 0]

ok.. aku post tadi sebab salah faham. oleh itu, semua sepatutnya sudah jelas bagi thread starter.