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.
Related
I have an static array, but when copying values over to a dynamic array, I get a bunch of nonsense padded on. I need for the resulting dynamic array to be exactly 8 characters
unsigned char cipherText[9]; //Null terminated
cout<<cipherText<<endl; //outputs = F,ÿi~█ó¡
unsigned char* bytes = new unsigned char[8]; //new dynamic array
//Loop copys each element from static to dynamic array.
for(int x = 0; x < 8; x++)
{
bytes[x] = cipherText[x];
}
cout<<bytes; //output: F,ÿi~█ó¡²²²²½½½½½½½½ε■ε■
You need to update your code to copy the null terminator:
unsigned char* bytes = new unsigned char[9]; //new dynamic array
//Loop copys each element from static to dynamic array.
for(int x = 0; x < 9; x++)
{
bytes[x] = cipherText[x];
}
cout<<bytes;
This is assuming that cipherText does in fact contain a null terminated string.
C-strings need to be null-terminated if you want to properly print it. The machine has to know where to stop printing.
To solve this, make bytes one larger (9 instead of 8) to make space for the null character and then append it to the end:
bytes[8] = '\0';
Now cout will now to only read the first 8 character from the array, after that it encounters the '\0' and will stop.
How do you know cipherText is null terminated? Simply defining it won't make it null terminated.
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.
I'm making a lexical analyzer and this is a function out of the whole thing. This function takes as argument a char, c, and appends this char to the end of an already defined char* array (yytext). It then increments the length of the text (yylen).
I keep getting segfaults on the shown line when it enters this function. What am I doing wrong here? Thanks.
BTW: can't use the strncpy/strcat, etc. (although if you want you can show me that implementation too)
This is my code:
extern char *yytext;
extern int *yylen;
void consume(char c){
int s = *yylen + 1; //gets yylen (length of yytext) and adds 1
//now seg faults here
char* newArray = new char[s];
for (int i = 0;i < s - 1;i++){
newArray[i] = yytext[i]; //copy all chars from existing yytext into newArray
}
newArray[s-1] = c; //append c to the end of newArray
for (int i = 0;i < s;i++){ //copy all chars + c back to yytext
yytext[i] = newArray[i];
}
yylen++;
}
You have
extern int *yylen;
but try to use it like so:
int s = (int)yylen + 1;
If the variable is an int *, use it like an int * and dereference to get the int. If it is supposed to be an int, then declare it as such.
That can t work:
int s = (int)yylen + 1; //gets yylen (length of yytext) and adds 1
char newArray[s];
use malloc or a big enought buffer
char * newarray=(char*)(malloc(s));
Every C-style string should be null-terminated. From your description it seems you need to append the character at c. So, you need 2 extra locations ( one is for appending the character and other for null-terminator ).
Next, yylen is of type int *. You need to dereference it to get the length (assuming it is pointing to valid memory location ). So, try -
int s = *yylen + 2;
I don't see the need of temporary array but there might be a reason why you are doing it. Now,
yytext[i] = newArray[i]; //seg faults here
you have to check if yytext is pointing to a valid write memory location. If yes, then is it long enough to fill the appending character plus null terminator.
But I would recommend using std::string than working with character arrays. Using it would be a one liner to solve the problem.
I wrote a very simple encryption program to practice c++ and i came across this weird behavior. When i convert my char* array to a string by setting the string equal to the array, then i get a wrong string, however when i create an empty string and add append the chars in the array individually, it creates the correct string. Could someone please explain why this is happening, i just started programming in c++ last week and i cannot figure out why this is not working.
Btw i checked online and these are apparently both valid ways of converting a char array to a string.
void expandPassword(string* pass)
{
int pHash = hashCode(pass);
int pLen = pass->size();
char* expPass = new char[264];
for (int i = 0; i < 264; i++)
{
expPass[i] = (*pass)[i % pLen] * (char) rand();
}
string str;
for (int i = 0; i < 264; i++)
{
str += expPass[i];// This creates the string version correctly
}
string str2 = expPass;// This creates much shorter string
cout <<str<<"\n--------------\n"<<str2<<"\n---------------\n";
delete[] expPass;
}
EDIT: I removed all of the zeros from the array and it did not change anything
When copying from char* to std::string, the assignment operator stops when it reaches the first NULL character. This points to a problem with your "encryption" which is causing embedded NULL characters.
This is one of the main reasons why encoding is used with encrypted data. After encryption, the resulting data should be encoded using Hex/base16 or base64 algorithms.
a c-string as what you are constructing is a series of characters ending with a \0 (zero) ascii value.
in the case of
expPass[i] = (*pass)[i % pLen] * (char) rand();
you may be inserting \0 into the array if the expression evaluates to 0, as well as you do not append a \0 at the end of the string either to assure it being a valid c-string.
when you do
string str2 = expPass;
it can very well be that the string gets shorter since it gets truncated when it finds a \0 somewhere in the string.
This is because str2 = expPass interprets expPass as a C-style string, meaning that a zero-valued ("null") byte '\0' indicates the end of the string. So, for example, this:
char p[2];
p[0] = 'a';
p[1] = '\0';
std::string s = p;
will cause s to have length 1, since p has only one nonzero byte before its terminating '\0'. But this:
char p[2];
p[0] = 'a';
p[1] = '\0';
std::string s;
s += p[0];
s += p[1];
will cause s to have length 2, because it explicitly adds both bytes to s. (A std::string, unlike a C-style string, can contain actual null bytes — though it's not always a good idea to take advantage of that.)
I guess the following line cuts your string:
expPass[i] = (*pass)[i % pLen] * (char) rand();
If rand() returns 0 you get a string terminator at position i.
I have a header file which contains a member variable declaration of a static char array:
class ABC
{
public:
static char newArray[4];
// other variables / functions
private:
void setArray(int i, char * ptr);
}
In the CPP file, I have the array initialized to NULL:
char ABC::newArray[4] = {0};
In the ABC constructor, I need to overwrite this value with a value constructed at runtime, such as the encoding of an integer:
ABC::ABC()
{
int i; //some int value defined at runtime
memset(newArray, 0, 4); // not sure if this is necessary
setArray(i,newArray);
}
...
void setArray(int i, char * value)
{
// encoding i to set value[0] ... value [3]
}
When I return from this function, and print the modified newArray value, it prints out many more characters than the 4 specified in the array declaration.
Any ideas why this is the case.
I just want to set the char array to 4 characters and nothing further.
Thanks...
How are you printing it? In C++ (and C), strings are terminated with a nul. (\0). If you're doing something like:
char arr[4] = {'u', 'h', 'o', 'h'};
std::cout << arr;
It's going to print "uhoh" along with anything else it runs across until it gets to a \0. You might want to do something like:
for (unsigned i = 0; i < 4; ++i)
std::cout << arr[i];
(Having a static tied to instances of a class doesn't really make sense, by the way. Also, you can just do = {}, though it's not needed since static variables are zero-initialized anyway. Lastly, no it doesn't make sense to memset something then rewrite the contents anyway.)
cout.write(arr, count_of(arr))
If count_of isn't defined in a system header:
template<typename T, size_t N>
inline size_t count_of(T (&array)[N]) { return N; }
Are you printing it using something like
printf("%s", newArray); //or:
cout << newArray;
? If so, you need to leave space for the nul-terminator at the end of the string. C strings are just arrays of characters, so there's no indication of the length of the string; standard library functions that deal with strings expect them to end in a nul (0) character to mark the ending, so they'll keep reading from memory until they find one. If your string needs to hold 4 characters, it needs to be 5 bytes wide so you can store the \0 in the fifth byte
You'll need a 5th character with a 0 byte to mark the end of the 4 character string, unless you use custom char-array output methods. If you set value[3] to something other than 0, you'll start printing bytes next to newArray in the static data area.
There's also no need to explicitly 0 initialize static data.
You can best catch those kinds of errors with valgrind's memcheck tool.
It is printing out a string that starts at the address &newArray[0] and ends at the first 0 in memory thereafter (called the null terminator).
char strArr[] = {"Hello"};
char strArr[] = {'H', 'e', "llo");
char strArr[] = "Hello";
char* strArr = "Hello"; // careful, this is a string literal, you can't mess with it (read-only usually)
...are all null terminated because anything in double quotes gets the null terminator tacked on at the end
char strArr[] = {'H', 'e', 'l', 'l', 'o'};
...is not null terminated, single quotes contain a single character and do not add a null terminator
Here are examples of adding a null terminator...
strArr[3] = '\0';
strArr[3] = NULL;
strArr[3] = 0;
With a bit loss of performance, you can fit into 4 byte.. in 'c-style'.
Print either 4 characters or until \0 is reached:
#include <cstdio>
#include <cstring>
...
//calculate length
size_t totalLength = sizeof(ABC::newArray) / sizeof(ABC::newArray[0]);
char* arrayEnd = (char*)memchr(ABC::newArray, '\0', totalLength);
size_t textLength = arrayEnd != 0 ?
arrayEnd-ABC::newArray : totalLength;
//print
fwrite(
ABC::newArray, //source array
sizeof(ABC::newArray[0]), //one item's size
textLength, //item count
stdout); //destination stream
By the way, try to use std::string and std::cout.