使用 C 将 char * 传递到 fopen
我正在编写一个程序，将数据从文件传递到数组中，但我在使用 fopen () 时遇到问题。当我将文件路径硬编码到参数中时（例如 fopen ("data/1.dat", "r");），它似乎工作正常，但是当我将它作为指针传递时，它会返回无效的。
请注意，如果从命令行输入，第 142 行将打印“data/1.dat”，因此 parse_args () 似乎正在工作。
132 int
133 main(int argc, char **argv)
134 {
135 FILE *in_file;
136 int *nextItem = (int *) malloc (sizeof (int));
137 set_t *dictionary;
138
139 /* Parse Arguments */
140 clo_t *iopts = parse_args(argc, argv);
141
142 printf ("INPUT FILE: %s.\n", iopts->input_file); /* This prints correct path */
143 /* Initialise dictionary */
144 dictionary = set_create (SET_INITAL_SIZE);
145
146 /* Use fscanf to read all data values into new set_t */
147 if ((in_file = fopen (iopts->input_file, "r")) == NULL)
148 {
149 printf ("File not found...\n");
150 return 0;
151 }
谢谢！ Rhys
更多：如果我在运行 set_create() (ln 144) 后尝试打印字符串，则该字符串不会打印。 （但是函数中根本没有任何对字符串的引用......）
47 set_t *
48 set_create(int size)
49 {
50 set_t *set;
51
52 /* set set_t members */
53 set->items = 0;
54 set->n_max = size;
55 set->lock = FALSE;
56
57 /* allocate memory for dictionary input */
58 set->data = (int *) malloc (size * sizeof (int));
59
60 return set;
61 }
如果我在 fopen() 之后调用这个函数，它确实可以工作。 不过，我看不出这对文件名有何影响...
再次感谢。
I'm writing a program that passes data from a file into an array, but I'm having trouble with fopen (). It seems to work fine when I hardcode the file path into the parameters (eg fopen ("data/1.dat", "r");) but when I pass it as a pointer, it returns NULL.
Note that line 142 will print "data/1.dat" if entered from command line so parse_args () appears to be working.
132 int
133 main(int argc, char **argv)
134 {
135 FILE *in_file;
136 int *nextItem = (int *) malloc (sizeof (int));
137 set_t *dictionary;
138
139 /* Parse Arguments */
140 clo_t *iopts = parse_args(argc, argv);
141
142 printf ("INPUT FILE: %s.\n", iopts->input_file); /* This prints correct path */
143 /* Initialise dictionary */
144 dictionary = set_create (SET_INITAL_SIZE);
145
146 /* Use fscanf to read all data values into new set_t */
147 if ((in_file = fopen (iopts->input_file, "r")) == NULL)
148 {
149 printf ("File not found...\n");
150 return 0;
151 }
Thanks!
Rhys
MORE: If I try to print the string after I run set_create() (ln 144), the string doesn't print. (But there isn't any reference to the string in the function at all...)
47 set_t *
48 set_create(int size)
49 {
50 set_t *set;
51
52 /* set set_t members */
53 set->items = 0;
54 set->n_max = size;
55 set->lock = FALSE;
56
57 /* allocate memory for dictionary input */
58 set->data = (int *) malloc (size * sizeof (int));
59
60 return set;
61 }
It does work if I call this function after fopen ().
I can't see how this is affecting the filename though...
Thanks again.
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评论（3）
您的新代码显示您正在写入无效内存。
set是一个指针，但您从未初始化它。您覆盖了一些随机内存，从而破坏了传递给fopen()的字符串指针。Your new code shows that you are writing to invalid memory.
setis a pointer but you never initialize it. You're overwriting some random memory and thereby destroying the pointer to the string that you're passing tofopen().您确定
parse_args工作正常吗？例如，如果它返回指向局部变量（或包含此类指针的结构）的指针，则像 iopts->input_file 这样的值很容易被后续函数调用破坏。Are you sure
parse_argsworks correctly? If it, for example, returns a pointer to a local variable (or a struct that contains such pointers), the values likeiopts->input_filewould easily be destroyed by subsequent function calls.第二部分是你的问题。设置未初始化。
澄清一下：您正在修改您无意的内容，导致 fopen() 失败。
That second part is your problem. set is not initialized.
To clarify: you're modifying stuff that you don't mean to, causing the fopen() to fail.