Dynamic fopen filepath in C

char filePath[200];   
printf("Enter filepath: \n");
fgets(filePath, 200, stdin);
f = fopen(filePath, "r");   

while(!feof(f)) // crashed on this line
{

}

I cannot for some reason get this to work. Please could some one point out what I am doing wrong on this.

Could you ad开发者_如何学JAVAvice the correct way to write code for opening a filepath specified by user through command prompt?

Thanks, Freddy

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fopen(3) returns NULL if it cannot open the file. You should always check for that. fgets(3) does that too, but your problem is probably the new-line character that it keeps in the returned string.

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fgets puts endline to the end of string (\r\n in Windows). So your filePath contains garbage at the end and file of that name doesn't exist. I would recommend to use scanf("%[^\r\n]s", filePath) instead. (scanf may differ on some implementations, please read your documentation.)

update: You should also ensure there won't be a buffer overflow by specifying buffer size. For example this way:

char filePath[100];
scanf("%99[^\r\n]s", filePath);

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Try something like:

size_t l = strlen(filePath);
if (filePath[l-1] == '\n') filePath[--l-1] = 0;
if (filePath[l-1] == '\r') filePath[--l-1] = 0;

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You check the error code from fgets and fopen. Both return null if they failed