0x00 And Char Arrays - C++ - Stack Overflow

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Learn more about Labs 0x00 and char arrays Ask Question Asked 12 years, 11 months ago Modified 12 years, 11 months ago Viewed 20k times 0

Why do char arrays stop right before a 0x00 byte is detected and how can this problem be avoided (perhaps by using another datatype (which one and why) or a "trick" with char)?

For example in the following code, the output is "a" only, the other bytes are not displayed:

unsigned char cbuffer[]={0x61,0x00,0x62,0x63,0x0}; std::string sbuffer=reinterpret_cast<const char*>(cbuffer); cout << sbuffer << endl;

Similarly in the following code, the output is "ab":

unsigned char cbuffer[]={0x61,0x62,0x00,0x63,0x0}; std::string sbuffer=reinterpret_cast<const char*>(cbuffer);

Straightforward and effective workarounds to the problem (where 0x00 is kept in the array as a normal byte) would be appreciated.

• c++
• char

Share Improve this question Follow asked Dec 14, 2011 at 19:09 Chris SmithChris Smith 76411 gold badge99 silver badges2222 bronze badges 3

• 4 Ummm. Because strings in C are arrays of char terminated with a zero byte? – Heath Hunnicutt Commented Dec 14, 2011 at 19:10
• std::vector with overloaded operator << (ostream&, const vector<char>&). – moshbear Commented Dec 14, 2011 at 19:16
• @HeathHunnicutt: Not necessarily. It's not uncommon to not use trailing zeros. The language itself only requires it for string literals. The libraries may also require it, but that's not the same thing. – Mooing Duck Commented Dec 14, 2011 at 19:18

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6 Answers 6

Sorted by: Reset to default Highest score (default) Trending (recent votes count more) Date modified (newest first) Date created (oldest first) 6

It's common in C to pass around strings as pointers to null-terminated char arrays. null is represented by 0x00. To make conversion easy, the std::string is constructable from a pointer to a null-terminated char array, which is what is happening with your code. But when it finds the null, it thinks that's the end of the string. If you cout a char array directly, you'll find it makes the same assumption, because they have no other way to determine the end of a string pointed to by a char*. (They could theoretically tell the length in your case, if they understood char (&)[], but almost nothing in the standard library does sadly).

The intended workarounds are to use this constructor instead:

int len = sizeof(cbuffer)/sizeof(cbuffer[0]); std::string sbuffer(cbuffer, len); //5 characters in cbuffer, 1 byte each

or

int len = sizeof(cbuffer)/sizeof(cbuffer[0]); std::cout.write(cbuffer, len); //5 characters in buffer, 1 byte each

However, you have to be careful with sizeof(cbuffer). If cbuffer is a char* (pointer) instead of a char(&)[] (array), then sizeof(ptr) will return the wrong value, and there is no way to get the correct length at that point, if the string is not null-terminated.

Share Improve this answer Follow edited Dec 14, 2011 at 21:29 answered Dec 14, 2011 at 19:15 Mooing DuckMooing Duck 66.6k1919 gold badges103103 silver badges165165 bronze badges 11

• 1 This is good except for not advocating usign sizeof to determine the buffer length. – Mark B Commented Dec 14, 2011 at 19:26
• @MarkB: I actually had that at first, but then removed it, because of the common "why does sizeof(ptr) return 4, I have 10 letters!". I'll edit the answer to describe the pros and cons. – Mooing Duck Commented Dec 14, 2011 at 19:27
• @curiousguy: What? Do you think my terminology/spelling is incorrect? – Mooing Duck Commented Dec 14, 2011 at 19:31
• @Mooing Duck. Your spelling is incorrect. NULL is a pointer and NUL is a character. – Heath Hunnicutt Commented Dec 14, 2011 at 19:38
• @HeathHunnicutt: Wikipedia disagrees: en.wikipedia.org/wiki/Null_character. NULL is the name of the character. It's commonly abbreviated as NUL however. – Mooing Duck Commented Dec 14, 2011 at 19:39

| Show 6 more comments 2

char arrays don't do anything

The C string functions use 0 to mark the end of a string. std::cout is overloaded for char arrays to print them as 'c' strings, if you want to print individual values you need to loop over the values, you might also want to output them as std::hex

In this case you are creating a std::String from a 'c' char array so the ctor of the std::string assumes that 'c' strings end at '0'. Since it's only passed an address in memory how else can it know where the string ends?

ps. If you want to store an array of bytes you should probably be using std::vector

Share Improve this answer Follow edited Dec 14, 2011 at 19:20 answered Dec 14, 2011 at 19:10 Martin BeckettMartin Beckett 96k2828 gold badges195195 silver badges268268 bronze badges 1

• he doesn't cout a char array, he couts a std::string object, which doesn't have this issue. – Mooing Duck Commented Dec 14, 2011 at 19:16

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Try:

#include <iostream> #include <string> int main() { unsigned char cbuffer[]={0x61,0x62,0x00,0x63,0x0}; // Here s1 is treating the cBuffer as a C-String // Thus it will only read upto the first '\0' character std::string s1(reinterpret_cast<const char*>(cbuffer)); std::cout << s1 << "\n"; // Here s2 is treating the cBuffer as an array. // It reads the specified length into the string. std::string s2(reinterpret_cast<const char*>(cbuffer), sizeof(cbuffer)/sizeof(cbuffer[0])); // Note: How std::cout prints the '\0' character may leave it empty. std::cout << s2 << "\n"; } Share Improve this answer Follow answered Dec 14, 2011 at 19:20 Loki AstariLoki Astari 264k8686 gold badges342342 silver badges570570 bronze badges Add a comment | 1

The 0x00 byte is used as a sentinel to mark the end of the string in C. The entire array however remains in memory. You can use an alternate constructor for std::string if you want the string to contain the entire character array. But printing that string would still give you only "ab". This decision to represent C strings in this manner is one of those arbitrary decisions that we are stuck with.

Share Improve this answer Follow answered Dec 14, 2011 at 19:20 shibumishibumi 37822 silver badges1111 bronze badges Add a comment | 0

0x00 is a non print character, 0..0x20, are all non print chars although some serve as line breaks. 0x00 serves to terminate a string.

Share Improve this answer Follow answered Dec 14, 2011 at 19:16 JohnJohn 6,63877 gold badges5353 silver badges8282 bronze badges Add a comment | 0

What do you want to be substituted (and printed) for 0x00 in the resulting string?

The constructor is responsible for conversion of char[] into a string. As others pointed out, you must use different constructor. The code below is working for me, but it is not very roboust. The first parameter must be a pointer to the array (you are free to use safer casting) and the second parameter is the length of the array (you are free to calculate this in a more sophisticated way).

#include <iostream> int main() { unsigned char cbuffer[]={0x61,0x00,0x62,0x63,0x00}; std::string sbuffer((char *)cbuffer,5); std::cout << sbuffer << std::endl; } Share Improve this answer Follow edited Dec 15, 2011 at 8:00 answered Dec 14, 2011 at 19:34 meolicmeolic 1,19722 gold badges2020 silver badges4848 bronze badges Add a comment |

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