我正在编写一个简单的程序来尝试实现strtok。只为我自己做一些练习
我想出了这个:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char * stringtok(char * str, char * delim){
static int last;
if(str != NULL){ last = 0; }
char * finder;
char * result = malloc(100);
while(delim[0] != '\0'){
if(finder = strchr(str+last, delim[0])){
memcpy(result, str + last, finder - str);
result[finder-str+1] = '\0';
last = finder - str;
return result;
}
delim++;
}
return NULL;
}
int main(){
char input[100]="This is my - message - hello can - you hear me?";
char input2[100] = "-";
char * pch;
pch = stringtok(input, input2);
while(pch != NULL){
printf("%s\n", pch);
pch = strtok(NULL, input2);
}
}
当我使用gcc -ggdb ./a.out时,我得到: “这是我的 ” 这很奇怪所以我通过GDB以断点22运行它,我得到了:
Breakpoint 1, main () at strtok.c:22
22 int main(){
(gdb) s
23 char input[100]="This is my - message - hello can - you hear me?";
(gdb)
29
(gdb)
stringtok (
str=0x7fffffffe040 "This is my - message - hello can - you hear me?",
delim=0x7fffffffdfd0 "-") at strtok.c:7
7 if(str != NULL){ last = 0; }
(gdb)
9 char * result = malloc(100);
(gdb)
10 while(delim[0] != '\0'){
(gdb)
11 if(finder = strchr(str+last, delim[0])){
(gdb)
12 memcpy(result, str + last, finder - str);
(gdb)
13 result[finder-str+1] = '\0';
(gdb)
14 last = finder - str;
(gdb)
15 return result;
(gdb)
19 return NULL;
(gdb)
main () at strtok.c:30
30 pch = stringtok(input, input2);
(gdb)
31 while(pch != NULL){
(gdb)
This is my
32 printf("%s\n", pch);
(gdb)
30 pch = stringtok(input, input2);
(gdb)
34 }
(gdb)
0x000000363d21ed5d in __libc_start_main () from /lib64/libc.so.6
(gdb)
Single stepping until exit from function __libc_start_main,
which has no line number information.
Program exited normally.
(gdb)
我觉得很奇怪,我有一个返回结果,然后它还会在之后返回NULL,我不确定为什么会发生这种情况。我只是读了gdb错了吗?它看起来像是从返回结果立即返回NULL。此外,看起来我的gdb正在输出1行的行,查看printf以及输出实际在哪里
答案 0 :(得分:0)
您的strtok函数无法正确处理str
NULL
。在第一次调用时,除了最后一个偏移量之外,还需要保存字符串。现在,当你第二次调用它时,你没有任何指向原始字符串的指针。