#include
INT main () {char S1 [9], S2 [9]; for (int i = 0; i <10; i ) {S1 [i] = 'a' i% 3; COUT <
<< "/ n" <
}
For (int J = 0; j <10; j )
{
S2 [J] = 'A' J / 3;
Cout <
<< "/ n" <
}
Return 0;
}
The array S1 is formed in a cyclic address A to C; ie: A, B, C, A, B, C, A, B, C ,. . .
The array S2 is formed by characters of A to C in three characters; ie: A, A, A, B, B, B, C, C, C ,. . .
This can be found by displaying a different 16-encycloped value by overflowing, you can find a spillation point.
The minimum value of the S1 error overflow address is: 0x42, only the constant loop of 0x41-0X43 this, so boldly speculative mantissa is 0x43-0x41 = 2; use S2 error overflow addresses: 0x43, each 3 digits, (So I am also confused) # @ @% ^ & * (() # $ @ () _ @ () # $
This is what the masters can do! Now I know it, the master is also found in this way, and I have to grow slowly, learn to learn the master! Hey! Learning, or learning. . .
