对于strcpy，sprintf，strcat这些不安全的函数的讨论应该算是不少了，大家都知道可以用strncpy，snprintf和strncat来替换它们。但实际工作中似乎这些“安全版”的函数带来的疑惑还是不小，总结了一下，列在这里。
1.sprintf(char * str, const char * format, ...) -> snprintf(char * str, size_t size, const char * format, ...)
vsprintf(char * str, const char * format, va_list ap) -> vsnprintf(char * str, size_t size, const char * format, va_list ap)
按照C/C++标准，snprintf和vsnprintf永远不会往str中输出多于size所指定的字节数的字符（包括结尾的'\0'），它们也保证了一定会在str中写入'\0'，所以在使用这个函数后不用担心结尾的问题。
举例:
char buf[5];
snprintf(buf, 5, "This is a test string."); // buf becomes "This", buf[4] is '\0'
snprintf(buf, 6, "This is a test string."); // ERROR: buffer overflow
snprintf(buf, 5, "abc"); // buf becomes "abc", the value of buf[3] is '\0', buf[4] is undefined.
然而，VC中的_snprintf函数并没有按照这样的规定来做，它在输出缓冲区不够大时就不会输出结尾的'\0'(跟strncpy的行为类似)。所以要让上面的程序工作正常，必须做相应的修改。
char buf[5];
_snprintf(buf, 5, "This is a test string.");// buf becomes "This ", buf[4] is ' '
buf[4] = 0; // buf[4] is '\0' now.
_snprintf(buf, 6, "This is a test string."); // ERROR: buffer overflow
_snprintf(buf, 5, "abc"); // buf becomes "abc", the value of buf[3] is '\0', buf[4] is undefined.
如果要保证可移植性，就得按VC的写法，做一次对于标准来说很多余的“填0”的操作，再通过定义宏来区别平台分别选用snprintf或_snprintf。
2. strcat(char * dest, const char * src) -> strncat(char * dest, const char * src, size_t n);
这个函数比较简单，它保证不会写入多于n+1个字符，并且保证最后一定以'\0'结束。
举例：
char dest[5] = "abc";
strncat(dest, "defghijklmn", 5 - 3 - 1); // dest becomes "abcd", dest[4] is '\0',
// always minus the buffer length by 1 as the value of n.
strncat(dest, "defghijklmn", 5 - 3); // ERROR: buffer overflow
3. strcpy(char * dest, const char * src) -> strncpy(char * dest, const char * src, size_t n);
strncpy是一个比较容易出错的函数，它保证复制src中不多于n字节的内容，但是如果src的前n个字节中没有包含'\0'，就会导致dest没有正常的以'\0'终止。另外，它还保证如果src的长度小于n，则dest剩余的部分都会以'\0'填充。在用这个函数时，有一个最佳实践就是先把dest以0填充，并传入dest的长度减1的值作为n，这样可以确保安全。
举例：
char buf[5] = {0}; // Always zero-fill the buffer,
// always use this form to initialize the stack arrays to get
//better performance over memset.
char * buf2 = new char[5];
memset(buf2, 0, 5); // Always zero-fill the buffer
strncpy(buf, "abcde", 5); // ERROR: buf is not null-terminated.
strncpy(buf2, "abcde", 5 - 1); // Right, always minus the buffer length by 1
//as the value of n. buf becomes "abcd", buf[4]
//is '\0' by initilization;
// This is a common error in the existing code.
char buf3[5];
const char * str = "Test";
strncpy(buf3, str, strlen(str)); // Wrong, buf may not null-terminated,
// potential buffer-overflow
strncpy(buf3, str, strlen(str)+1); // Wrong, potential buffer-overflow.
//No difference with using strcpy.
