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Home/ Questions/Q 8429497
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Editorial Team
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Editorial Team
Asked: June 10, 20262026-06-10T05:21:21+00:00 2026-06-10T05:21:21+00:00

I am using snprintf to concatenate a string to a char array: char buf[20]

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I am using snprintf to concatenate a string to a char array:

char buf[20] = "";
snprintf(buf, sizeof buf, "%s%s", buf, "foo");
printf("%s\n", buf);
snprintf(buf, sizeof buf, "%s%s", buf, " bar");
printf("%s\n", buf);

The problem is the second concatenation to buf instead of adding "bar", replaces "foo" with it. The output is like:

foo
bar

The first %s should keep buf (which in this case holds "foo") there. And the second %s should attach "bar" to it. Right?

What am I doing wrong?

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  1. Editorial Team
    Editorial Team
    2026-06-10T05:21:22+00:00Added an answer on June 10, 2026 at 5:21 am

    You’re violating the restrict contract on snprintf, which states that no other argument can overlap the buffer.

    Copying the input into itself is a waste of effort anyway. snprintf returns the number of characters which formatting would require, so take advantage of this for appending:

    char buf[20] = "";
    char *cur = buf, * const end = buf + sizeof buf;
    cur += snprintf(cur, end-cur, "%s", "foo");
    printf("%s\n", buf);
    if (cur < end) {
        cur += snprintf(cur, end-cur, "%s", " bar");
    }
    printf("%s\n", buf);
    
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