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copied!<p>The issue here is that the length of the string <code>a</code> cannot be easily determined. For example, your code..</p> <pre><code>char *a = "\0hey\0\0"; </code></pre> <p>.. allocates seven bytes to the string, the last being the NULL terminator. Using a function like <code>strlen</code> would return 0.</p> <p>If you know the precise length of the string, then you can write or iterate over the bytes thus:</p> <pre><code>#ifdef ESCAPE_NULLS int i; for (i = 0; i <= 6; i++) if (a[i] == 0) printf("\\0"); else printf("%c", a[i]); #else write(1, a, 6); #endif </code></pre> <p>But you have to know about the 6.</p> <p>The alternative is not to use NULL-terminated strings, and instead implement an alternative storage mechanism for your bytes; for example a length-encoded array.</p> <pre><code>#include <stdio.h> typedef struct { int length; char *bytes; } bytearr; void my_printf(bytearr *arr) { #ifdef ESCAPE_NULLS int i; for (i = 0; i <= arr->length; i++) if (arr->bytes[i] == 0) printf("\\0"); else printf("%c", arr->bytes[i]); #else write(1, arr->bytes, arr->length); #endif } void main(void) { bytearr foo = { 6, "\0hey\0\0" }; my_printf(&foo); } </code></pre> <p>Graceless, but hopefully you get the idea.</p> <p><strong>Edit: 2011-05-31</strong></p> <p>Rereading the question I just noticed the word "concatenate". If the NULL characters are to be copied faithfully from one place in memory to another (not backslash-escape), and you know the total number of bytes in each array beforehand, then you can simply use <a href="http://linux.die.net/man/3/memcpy" rel="noreferrer"><code>memcpy</code></a>.</p> <pre><code>#include <string.h> char *a = "\0hey\0\0"; /* 6 */ char *b = "word\0up yo"; /* 10 */ char *c = "\0\0\0\0"; /* 4 */ void main(void) { char z[20]; char *zp = z; zp = memcpy(zp, a, 6); zp = memcpy(zp, b, 10); zp = memcpy(zp, c, 4); /* now z contains all 20 bytes, including 8 NULLs */ } </code></pre>

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