strcpy not copying the full string

Question

I'm using Visual C++ 6.0, and I have the code below.

char filename[1000], string[5]; FILE *fin; strcpy(filename, argv[3]); if ((fin = fopen(filename, "r")) != NULL) { fgets(string, 100, fin); string[strlen(string)-1] = NULL; printf("filename = %s\n", filename); printf("argv[3]= %s\n", argv[3]); printf("string = %s\n", string); }

argv[3] is the full path and filename, e.g. C:\Users\Desktop\file.txt, and the content of the file is

1 2 3

So "1" should be stored in the "string" variable.

However, for about 1 out of 4 runs, I would get the output

filename = C:\Users\Desktop\file.tx argv[3] = C:\Users\Desktop\file.txt string = <very long garbage value>

Why did

strcpy(filename, argv[3]);

not copy the entire string, missing the last "t"? And why is fin not NULL in this case, since the file should not have existed?

I should also add that this code exists in a multi-thread program, but only 1 thread executes this code.

If you are using C++, why are you still using char [] and FILE instead of std::string and streams? — user1233508
– user1233508, Commented Nov 15, 2012 at 10:46
I'm more familiar with char[] and FILE. Would this have made a difference? — Rayne
– Rayne, Commented Nov 16, 2012 at 0:55
Yes, it would have. std::string, like most other std containers, expands automagically to contain all the data you put into it, saving you from having to track lengths and memory allocation manually. — user1233508
– user1233508, Commented Nov 16, 2012 at 5:01

jcoder · Accepted Answer · 2012-11-15 10:43:25Z

0

string[5];

You have only allocated enough space for 4 characters and a null terminator but your fgets is reading up to 100.

fgets(string, 100, fin);

answered Nov 15, 2012 at 10:43

jcoder

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1 Comment

Rayne Over a year ago

Yes, but wouldn't fgets also read up to the \n, which the file has after the "1". Anyway, this line comes after the strcpy, which I believe is the main problem for the garbage value I'm getting for "string".

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