I always get a garbage value like this 'Íýýýý««««««««îþîþ' at the end when i output my array. What am I doing wrong?
void func()
{
const int size = 100;
char * buffer = new char[size];
for(int i=0; i<size; i++)
buffer[i] = ' ';
cout<<buffer;
}
However if I use a for loop to output the buffer, there is no garbage value.
Because you don't null terminate your buffer, std::cout.operator<<(char*) will try to find \0 as its terminating character.
As pointed out in comments, feel free to append that \0 to the end of your buffer :).
ScarletAmaranth is right. C style strings (an array of char) must finish with the char '\0'. That's the way functions that play with char arrays (cout in this case) know when the string finish in memory. If you forget the '\0' character at the end of the array, cout prints all the chars in the array and then goes on printing any data in memory after the array. These other data is rubbish, and that's why you see these strange characters.
If you use the string type (more C++ style) instead of char arrays, you won't have this problem because string type don't use '\0' as a string delimiter.
On the other hand, you don't see rubbish when use a loop to iterate over the 100 elements of the array just because you only print these 100 chars. I mean, you control exactly what you are printing and know when to stop, instead of leaving the cout function figure out when to stop printing.
Related
I am new to c++ and am still figuring out file streams. I am trying to put a character array into a file that I will be viewing with a hex editor.
I have done different strings, but whenever I put in a null byte, the file ends.
ofstream stream;
char charArray[256];
for (int i = 0; i <= 255; i++)charArray[i] = i;
stream.open("C:\\testfile.myimg");
if (!stream.is_open()) exit(1);
stream << charArray << endl;
return 0;
I want to output bytes with ascending values, but if I start with the null byte, then c++ thinks the character array ends before it starts
Instead of:
stream << charArray << endl;
use:
stream.write(charArray, sizeof(charArray));
stream.write("\n", 1); // this is the newline part of std::endl
stream.flush(); // this is the flush part of std::endl
The first one assumes that you are sending a null-terminated string (because you're passing a char* - see the end of the answer why). That's why when the code encounters the very first char with value 0, which is '\0' (the null-terminator), it stops.
On the other hand, the second approach uses an unformatted output write, which will not care about the values inside charArray - it will take it (as a pointer to its first element) and write sizeof(charArray) bytes starting from that pointer to the stream. This is safe since it's guaranteed that sizeof(char) == 1, thus sizeof(charArray) will yield 256.
What you need to consider is array decaying. It will work in this case (the sizeof thing), but it will not work if you simply pass a decayed pointer to the array's first element. Read more here: what's array decaying?
I have a string like,
string str="aaa\0bbb";
and I want to copy the value of this string to a char* variable. I tried the following methods but none of them worked.
char *c=new char[7];
memcpy(c,&str[0],7); // c="aaa"
memcpy(c,str.data(),7); // c="aaa"
strcpy(c,str.data()); // c="aaa"
str.copy(c,7); // c="aaa"
How can I copy that string to a char* variable without loosing any data?.
You can do it the following way
#include <iostream>
#include <string>
#include <cstring>
int main()
{
std::string s( "aaa\0bbb", 7 );
char *p = new char[s.size() + 1];
std::memcpy( p, s.c_str(), s.size() );
p[s.size()] = '\0';
size_t n = std::strlen( p );
std::cout << p << std::endl;
std::cout << p + n + 1 << std::endl;
}
The program output is
aaa
bbb
You need to keep somewhere in the program the allocated memory size for the character array equal to s.size() + 1.
If there is no need to keep the "second part" of the object as a string then you may allocate memory of the size s.size() and not append it with the terminating zero.
In fact these methods used by you
memcpy(c,&str[0],7); // c="aaa"
memcpy(c,str.data(),7); // c="aaa"
str.copy(c,7); // c="aaa"
are correct. They copy exactly 7 characters provided that you are not going to append the resulted array with the terminating zero. The problem is that you are trying to output the resulted character array as a string and the used operators output only the characters before the embedded zero character.
Your string consists of 3 characters. You may try to use
using namespace std::literals;
string str="aaa\0bbb"s;
to create string with \0 inside, it will consist of 7 characters
It's still won't help if you will use it as c-string ((const) char*). c-strings can't contain zero character.
There are two things to consider: (1) make sure that str already contains the complete literal (the constructor taking only a char* parameter might truncate at the string terminator char). (2) Provided that str actually contains the complete literal, statement memcpy(c,str.data(),7) should work. The only thing then is how you "view" the result, because if you pass c to printf or cout, then they will stop printing once the first string terminating character is reached.
So: To make sure that your string literal "aaa\0bbb" gets completely copied into str, use std::string str("aaa\0bbb",7); Then, try to print the contents of c in a loop, for example:
std::string str("aaa\0bbb",7);
const char *c = str.data();
for (int i=0; i<7; i++) {
printf("%c", c[i] ? c[i] : '0');
}
You already did (not really, see edit below). The problem however, is that whatever you are using to print the string (printf?), is using the c string convention of ending strings with a '\0'. So it starts reading your data, but when it gets to the 0 it will assume it is done (because it has no other way).
If you want to simply write the buffer to the output, you will have to do this with something like
write(stdout, c, 7);
Now write has information about where the data ends, so it can write all of it.
Note however that your terminal cannot really show a \0 character, so it might show some weird symbol or nothing at all. If you are on linux you can pipe into hexdump to see what the binary output is.
EDIT:
Just realized, that your string also initalizes from const char* by reading until the zero. So you will also have to use a constructor to tell it to read past the zero:
std::string("data\0afterzero", 14);
(there are prettier solutions probably)
Why does this code produce runtime issues:
char stuff[100];
strcat(stuff,"hi ");
strcat(stuff,"there");
but this doesn't?
char stuff[100];
strcpy(stuff,"hi ");
strcat(stuff,"there");
strcat will look for the null-terminator, interpret that as the end of the string, and append the new text there, overwriting the null-terminator in the process, and writing a new null-terminator at the end of the concatenation.
char stuff[100]; // 'stuff' is uninitialized
Where is the null terminator? stuff is uninitialized, so it might start with NUL, or it might not have NUL anywhere within it.
In C++, you can do this:
char stuff[100] = {}; // 'stuff' is initialized to all zeroes
Now you can do strcat, because the first character of 'stuff' is the null-terminator, so it will append to the right place.
In C, you still need to initialize 'stuff', which can be done a couple of ways:
char stuff[100]; // not initialized
stuff[0] = '\0'; // first character is now the null terminator,
// so 'stuff' is effectively ""
strcpy(stuff, "hi "); // this initializes 'stuff' if it's not already.
In the first case, stuff contains garbage. strcat requires both the destination and the source to contain proper null-terminated strings.
strcat(stuff, "hi ");
will scan stuff for a terminating '\0' character, where it will start copying "hi ". If it doesn't find it, it will run off the end of the array, and arbitrarily bad things can happen (i.e., the behavior is undefined).
One way to avoid the problem is like this:
char stuff[100];
stuff[0] = '\0'; /* ensures stuff contains a valid string */
strcat(stuff, "hi ");
strcat(stuff, "there");
Or you can initialize stuff to an empty string:
char stuff[100] = "";
which will fill all 100 bytes of stuff with zeros (the increased clarity is probably worth any minor performance issue).
Because stuff is uninitialized before the call to strcpy. After the declaration stuff isn't an empty string, it is uninitialized data.
strcat appends data to the end of a string - that is it finds the null terminator in the string and adds characters after that. An uninitialized string isn't gauranteed to have a null terminator so strcat is likely to crash.
If there were to intialize stuff as below you could perform the strcat's:
char stuff[100] = "";
strcat(stuff,"hi ");
strcat(stuff,"there");
Strcat append a string to existing string. If the string array is empty, it is not going go find end of string ('\0') and it will cause run time error.
According to Linux man page, simple strcat is implemented this way:
char*
strncat(char *dest, const char *src, size_t n)
{
size_t dest_len = strlen(dest);
size_t i;
for (i = 0 ; i < n && src[i] != '\0' ; i++)
dest[dest_len + i] = src[i];
dest[dest_len + i] = '\0';
return dest;
}
As you can see in this implementation, strlen(dest) will not return correct string length unless dest is initialized to correct c string values. You may get lucky to have an array with the first value of zero at char stuff[100]; , but you should not rely on it.
Also, I would advise against using strcpy or strcat as they can lead to some unintended problems.
Use strncpy and strncat, as they help prevent buffer overflows.
I need an empty char array, but when i try do thing like this:
char *c;
c = new char [m];
int i;
for (i = 0; i < m; i++)
c[i] = 65 + i;
and then I print c. can see that c = 0x00384900 "НННННННээээ««««««««юоюою"
after cycle it becomes: 0x00384900 "ABCDEFGээээ««««««««юоюою"
How can I solve this problem? Or maybe there is way with string?
If you're trying to create a string, you need to make sure that the character sequence is terminated with the null character \0.
In other words:
char *c;
c = new char [m+1];
int i;
for (i = 0; i < m; i++)
c[i] = 65 + i;
c[m] = '\0';
Without it, functions on strings like printf won't know where the string ends.
printf("%s\n",c); // should work now
If you create a heap array, OS will not initialiase it.
To do so you hvae these options:
Allocate an array statically or globally. The array will be filled with zeroes automatically.
Use ::memset( c, 0, m ); on heap-initialised or stack array to fill it with zeroes.
Use high-level types like std::string.
I believe that's your debugger trying to interpret the string. When using a char array to represent a string in C or C++, you need to include a null byte at the end of the string. So, if you allocate m + 1 characters for c, and then set c[m] = '\0', your debugger should give you the value you are expecting.
If you want a dynamically-allocated string, then the best option is to use the string class from the standard library:
#include <string>
std::string s;
for (i = 0; i < m; i++)
s.push_back(65 + i);
C strings are null terminated. That means that the last character must be a null character ('\0' or just 0).
The functions that manipulate your string use the characters between the beginning of the array (that you passed as parameter, first position in the array) and a null value. If there is no null character in your array the function will iterate pass it's memory until it finds one (memory leak). That's why you got some garbage printed in your example.
When you see a literal constant in your code, like printf("Hello");, it is translate into an array of char of length 6 ('H', 'e', 'l', 'l', 'o' and '\0');
Of course, to avoid such complexity you can use std::string.
NoobQuestion:
I heard that filling a char array can be terminated early with the null char. How is this done?
I've searched every single google result out there but still have come up empty handed.
Do you mean something like this:
char test[11] = "helloworld";
std::cout << test << std::endl;
test[2] = 0;
std::cout << test;
This outputs
helloworld
he
?
That's a convention called "null-terminated string". If you have a block of memory which you treat as a char buffer and there's a null character within that buffer then the null-terminated string is whatever is contained starting with the beginning of the buffer and up to and including the null character.
const int bufferLength = 256;
char buffer[bufferLength] = "somestring"; //10 character plus a null character put by the compiler - total 11 characters
here the compiler will place a null character after the "somestring" (it does so even if you don't ask to). So even though the buffer is of length 256 all the functions that work with null-terminated strings (like strlen()) will not read beyond the null character at position 10.
That is the "early termination" - whatever data is in the buffer beyond the null character it is ignored by any code designed to work with null-terminated strings. The last part is important - code could easily ignore the null character and then no "termination" would happen on null character.