Today's entire course has only such questions, but there are many things that have been learned. Here are this topic:
Character numbers are converted to integer values (characters can be arbitrary: such as "342a") encountered other no numbers to take the front number.
The procedures I have written are as follows, I think it is good:
#define n 10;
INT CATIO (const char * str) / * const is constant, so the address here will not return to the inactive parameters * /
{
Int num [n];
INT i = 0; j = 1, n = 0;
For (; * str ; i )
{
IF (* STR <48 || * STR> 57)
Break; / * Judging whether the numeric value * /
NUM [I] = * STR-48;
}
For (i- = 1; i> = 0; I -)
{
N = NUM [i] * j;
J * = 10;
}
Return n;
}
Do you say that it is more simple? Now I can't see another program of reading the following, I will conclude. as follows:
Long catio (char C []);
{
INT N, D;
Char * q, * p;
Long E = 1, s = 0;
For (q = p = c, n = 0; * p! = '/ 0' && * p> = '0' && * P <= '9'; p , n , e * = 10);
While (n> 0)
{
D = * q ;
Switch (d)
{
Case 48: D = 0; Break; / * too long, slightly * /
:
:
Case 57: D = 9; Break;
}
S = D * (e / = 10);
n--;
}
Return (s);
}
Now let's take a look, but although this procedure is complicated than me, there is also his ideas and ideas. Like that for loop, a command has been very convenient. In fact, we can continue to transform this procedure, we follow the teacher's ideas to evolve it step by step, now look at the following: long catio (char c []); {INT N, D; char * q, * p; long e = 1, S = 0; for (q = p = c, n = 0; * p && * p> = '0' && * p <= '9'; p , n , e * = 10);
While (n> 0) {d = * q - '0'; s = D * (E / = 10); n--;} return (s);}
