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自從我在C中寫下我的最後一行以來已經有很多年了，現在我正在嘗試再次開始創建一些函數並將其用作php的擴展。如何在c/gcc中返回腳本？
這是一個簡單的函數來創建「小網址」
讓說：
a0bg a0bf a0bh
我遇到的問題是，當我以「增量」的像ZZZ字符串，然後我得到：總線錯誤
否則，如果我增加ABR比如我得到的結果是：ABS
在某些時候我覺得我的問題這裏返回字符串結果
兩個代碼是全功能的，因爲我張貼他們在這裏。
編譯我使用：的gcc -o append_id_test.c append_id
我在OS X Leopard的
我這樣做的時候，它的工作原理：（請注意在電話會議上在功能append_id）
#include <stdlib.h>
#include <string.h>
#include <stdio.h>
char* append_id(char*);
int main(void) {
char *x;
char *incremented;
x = "aaab";
printf("%s\n", append_id(x)); // should print 00000 (lenght 5)
incremented = (char *) malloc((strlen(x) + 2) * sizeof(char));
incremented = append_id(x);
printf("---> %s\n", incremented); // should print 00000 (lenght 5)
}
char* append_id(char *id) {
int x;
char* new_id;
int id_size = strlen(id);
new_id = (char *) malloc((strlen(id) + 2) * sizeof(char));
for (x = 0; x < id_size; x++)
{
new_id[x] = '0';
}
strcat(new_id, "0");
return new_id;
}
但整個代碼不工作（正如你所看到的調用append_id功能是在相同的方式進行上面的例子）
#include <stdlib.h>
#include <string.h>
#include <stdio.h>
char* append_id(char*);
char* increment_id(char*, int);
char* get_next_id(char*);
int main(int argc, char *argv[]) {
char *x;
int a;
x = "zz";
printf("incrementando %s -> %s\n", "zz", get_next_id(x));
return 0;
}
char * get_next_id(char *last_id)
{
int x, pos;
char *next_id;
char is_alnum = 1;
// if the last id is -1 (non-existant), start at the begining with 0
if (strlen(last_id) == 0)
{
next_id = "0";
}
else
{
// check the input
for(x = 0; last_id[x]; x++)
{
if(!isalnum(last_id[x]))
{
is_alnum = 0;
break;
}
}
if (is_alnum == 0)
{
return "";
}
// all chars to lowercase
for(x = 0; last_id[x]; x++)
{
last_id[x] = tolower(last_id[x]);
}
// loop through the id string until we find a character to increment
for (x = 1; x <= strlen(last_id); x++)
{
pos = strlen(last_id) - x;
if (last_id[pos] != 'z')
{
next_id = increment_id(last_id, pos);
break; // <- kill the for loop once we've found our char
}
}
// if every character was already at its max value (z),
// append another character to the string
if (strlen(next_id) == 0)
{
next_id = (char *) malloc((strlen(last_id) + 2) * sizeof(char));
next_id = append_id(last_id);
}
}
return next_id;
}
char* append_id(char *id) {
int x;
char* new_id;
int id_size = strlen(id);
new_id = (char *) malloc((strlen(id) + 2) * sizeof(char));
for (x = 0; x < id_size; x++)
{
new_id[x] = '0';
}
strcat(new_id, "0");
return new_id;
}
char* increment_id(char *id, int pos){
char current, new_char;
char * new_id ;
int x;
new_id = (char *) malloc((strlen(id) + 1) * sizeof(char));
current = id[pos];
if (current >= 0x30 && current <= 0x39)
{
if (current < 0x39)
{
new_char = current + 1;
}
else // if we're at 9, it's time to move to the alphabet
{
new_char = 'a';
}
}
else // move it up the alphabet
{
new_char = current + 1;
}
for (x = 0; x < strlen(id); x++)
{
if (x == pos) {
new_id[x] = new_char;
}
else {
new_id[x] = id[x];
}
}
// set all characters after the one we're modifying to 0
if (pos != (strlen(new_id) - 1))
{
for (x = (pos + 1); x < strlen(new_id); x++)
{
new_id[x] = '0';
}
}
return new_id;
}
我看到malloc的，但我沒有看到免費的... – fortran 2009-07-14 14:50:40
說：「我的代碼的這一小部分工作，但與其他2個功能，是4倍大的其他部分不」，是不是很有幫助。 – dborba 2009-07-14 15:02:32
我只是想向您展示問題的背景：我認爲（我不確定，是從第二部分的第74行開始的問題） – 2009-07-14 15:09:03