例如,我有一个字符串的ptr并将ptr移动到字符串中的最后一个字符并使用*p--向后迭代到字符串的开头,并且我在数组开始之前迭代到位置1这样可以吗？或者我会获得访问冲突？我只是移动指针 - 不访问.它似乎在我的代码中工作,所以想知道它是不是很糟糕的做法？
这是一个带有*next-- = rem +'A'的样本行; 是一个我在质疑如果好吗???
#include <stdio.h> /* printf */
#include <string.h> /* strlen, strcpy */
#include <stdlib.h> /* malloc/free */
#include <math.h> /* pow */
/* AAAAA (or whatever length) = 0, to ZZZZZ. base 26 numbering system */
static void getNextString(const char* prev, char* next) {
int count = 0;
char tmpch = 0;
int length = strlen(prev);
int i = 0;
while((tmpch = *prev++) != 0) {
count += (tmpch - 'A') * (int)pow(26.0, length - i - 1);
++i;
}
/* assume all strings are uppercase eg AAAAA */
++count;
/*if count above ZZZ... then reset to AAA... */
if( count >= (int)pow(26.0, length))
count = 0;
next += (length-1); /* seek to last char in string */
while(i-- > 0) {
int rem = count % 26;
count /= 26;
*next-- = rem + 'A'; /*pntr positioned on 1 before array on last iteration - is OK? */
}
}
int main(int argc, char* argv[])
{
int buffsize = 5;
char* buff = (char*)malloc(buffsize+1);
strcpy(buff, "AAAAA");
int iterations = 100;
while(--iterations){
getNextString(buff, buff);
printf("iteration: %d buffer: %s\n", iterations, buff);
}
free(buff);
return 0;
}