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fscanf / fscanf_s difference in behaviour

Asked 2010-06-24T18:31:13.967
10

I'm puzzled by the following difference in behaviour:

// suppose myfile.txt contains a single line with the single character 's'
    errno_t res;
    FILE* fp;
    char cmd[81];

    res = fopen_s(&fp, "D:\\myfile.txt", "rb" );
    fscanf(fp,"%80s",cmd); // cmd now contains 's/0'
    fclose(fp);

    res = fopen_s(&fp, "D:\\myfile.txt", "rb" );
    fscanf_s(fp,"%80s",cmd); // cmd now contains '/0' !
    fclose(fp);

The results do not depend in the order of call (i.e., call fscanf_s first, you'd get the empty string first). Compiled on VC++ - VS2005. Can anyone reproduce? Can anyone explain?

Thanks!

c++ visual-c++ visual-studio-2005 crt
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1 Answer

10

From the docs on fscanf_s(), http://msdn.microsoft.com/en-us/library/6ybhk9kc.aspx:

The main difference between the secure functions (with the _s suffix) and the older functions is that the secure functions require the size of each c, C, s, S and [ type field to be passed as an argument immediately following the variable. For more information, see scanf_s, _scanf_s_l, wscanf_s, _wscanf_s_l and scanf Width Specification.

And http://msdn.microsoft.com/en-us/library/w40768et.aspx:

Unlike scanf and wscanf, scanf_s and wscanf_s require the buffer size to be specified for all input parameters of type c, C, s, S, or [. The buffer size is passed as an additional parameter immediately following the pointer to the buffer or variable. For example, if reading a string, the buffer size for that string is passed as follows:

char s[10];

scanf("%9s", s, 10);

So you should call it like so:

fscanf_s(fp,"%80s",cmd, sizeof(cmd));
answered 2010-06-24T18:46:37.237

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