c - How will 'fopen' interpret a 'char[]' whose capacity is larger than the string it contains? -

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i writing program can encrypt text file.

first need ask users file name. did like:

char file_name[50]; fgets(file_name, 50, stdin);

but didn't work. how can this?

i confused if store file name in char array has, say, 50 elements, file name has 10 characters. when pass array or pointer fopen, program going remaining 40 elements of array? storing value? passed fopen?

in c, string \0 terminated sequence of characters. long there \0 within 50 bytes of file_name, file_name contains valid string fopen().

the reason fopen() failed name passed had \n in after reading input. because fgets() stores newline character buffer. have remove before using string file name.

char *p = strrchr(file_name, '\n'); if (p) *p = '\0';

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