C - Opening a file inside a function using fopen

问题

I am using Visual Studio 2005 (C\C++).

I am passing a string into a function as a char array. I want to open the file passed in as a parameter and use it. I know my code works to an extent, because if I hardcode the filename as the first parameter it works perfectly.

I do notice if I look at the value as a watch, the value includes the address aside the string literal. I have tried passing in the filename as a pointer, but it then complains about type conversion with __w64. As I said before it works fine with "filename.txt" in place of fileName. I am stumped.

void read(char fileName[50],int destArray[MAX_R][MAX_C],int demSize[2])
{
    int rows=0;
    int cols=0;
    int row=0;
    int col=0;
    FILE * f = fopen(fileName,"r");
...

The calling function code is:

char in_filename[50];
int dem[MAX_R][MAX_C]; 
int dem_size[2];
get_user_input( in_filename);
read(in_filename, dem, dem_size );

In the watch I added for filename the correct text appears, so the data is getting passed in.

回答1:

If you're using fopen() then you're coding in C, not C++. Also, this is not how you pass arrays to functions. The syntax for the parameter list is

void f(char arr[], unsigned int arr_size);

In the case of multidimensional arrays you must specify the size of the right-most dimension explicitly:

void f(char arr[][20], unsigned int arr_size);

That said, try changing the parameter from char fileName[50] to char* fileName.

来源：https://stackoverflow.com/questions/2755663/c-opening-a-file-inside-a-function-using-fopen

标签

fopen