试图打开文件名可变的文件时，c程序崩溃

Question

As stated in the title, i ask for the user to provide the filename and i use gets to save it in str . Then i try to access the file using the name and the program crashes.

int openFile(FILE *fp){
    puts("What's the name of the file (and format) to be accessed?");
    char str[64];
    gets(str);  
    fp = fopen((const char *)str, 'r');
    ...
    return 0;

In main:

FILE *fp; // file pointer

openFile(fp);

The filename i enter (data.txt) is indeed in the same directory as the rest of the project so that should not be the problem. I've tried testing if the file is opened correctly (which it should) but it keeps crashing right after i give the name.

Answer 1

The main problem is that you are trying to set an argument passed by value in a function and expect the value to be changed outside. This can't work.

Currently you have:

void openFile(FILE* fp) {
  fp = ...
}

int main()
{
  FILE* fp;
  openFile(fp);
}

But fp in main() is passed as a pointer by value. Which means that inside openFile you are setting a local variable, while the passed one is not modified.

To solve the problem you can:

• directly return a FILE* from openFile
• accept a pointer to pointer argument to be able to set it, eg void openFile(FILE** fp) and then openFile(&fp)

Mind that the second argument of fopen is a const char* not a single char , "r" should be used.

Answer 2

It should be fp = fopen(str, "r"); , because fopen() expects mode as a char * pointing to a string, rather than a single char .

Also, since parameters in C are passed by value , your fp won't get modified after openFile() is called. To get it work, you'll have to rewrite it, and call it by openFile(&fp); . Here is an example:

void openFile(FILE **fp) {
    puts("What's the name of the file (and format) to be accessed?");
    char str[64];
    fgets(str, 64, stdin);
    str[strcspn(str, "\n")] = '\0';  
    *fp = fopen(str, "r");
}

fgets() is used to provide buffer overflow protection.