I have a const char* variable which takes values from a function that I have wrote.
When I write this variable to a file many times it writes nothing. So it must be empty or filled in with space.The strange thing is that in the txt file that I write it changes line every time, when it has value or not.Why is that?Does it mean that the returned value from the function has a \n?
how can I check if a value of a const char * is empty or in general how can I check character by character the value in char*?
Since C/C++ pointers can be interpreted as arrays of values the pointers point to, the two ways of checking values of a char* is by applying an indexing operator or by using pointer arithmetics. You can do this:
const char *p = myFunctionReturningConstChar();
for (int i = 0 ; p[i] ; i++) {
if (p[i] == '\n') printf("New line\n");
}
or this:
const char *p = myFunctionReturningConstChar();
while (*p) {
if (*p == '\n') printf("New line\n");
p++;
}
In addition, C++ library provides multiple functions for working with C strings. You may find strlen helpful to check if your pointer points to an empty string.
Related
#include <stdio.h>
char strA[80] = "A string to be used for demonstration purposes";
char strB[80];
char *my_strcpy(char *destination, char *source)
{
char *p = destination;
while (*source != '\0')
{
*p++ = *source++;
}
*p = '\0';
return destination;
}
int main(void)
{
my_strcpy(strB, strA);
puts(strB);
}
so my question here is that when i take out the portion:
//*p= '\0';
it prints the exact same answer, so why is this necessary? from my understanding, \0 is a nul portion of memory after a string but since the array strA already contains the nul portion since its in "" is it really necessary?
It seems you already know the importance of the null terminator, but the point is, you defined char strB[80]; in external namespace (with static life span), which causes initialization of the array strB, which sets all bytes of it to zero. That's why you can't observe the difference （because even if you don't append a null character, the rest of strB already is).
Moving the definition of strB makes this visible. strA doesn't need moving because it doesn't matter.
In actuality, this code
while (*source != '\0')
{
*p++ = *source++;
}
// *p = '\0';
When *source reaches a null character, it's not copied to *p, so you need to manualky add a terminator for that.
Your loop stops when it sees the \0 and so it is not copied to the destination and the destination is not NUL terminated. Is that a problem?
Not if your destination buffer is initialized to all 0s
Not if your code is willing to deal with fixed length strings (so the my_strcpy signature would need to change to return the length)
In general YES - the 0 terminated C string is such a common thing that not following the convention is asking for trouble,
Whether you 0 terminate or not the rest of the values will be the same as they were when you started. The 0 termination just makes your character array a "standard C string".
For arguments sake: Assuming you knew every string had space for 80 chars you could just do
for(int i = 0; i < 80; i++)
{
dest[i] = src[i];
}
The effect is the same and assuming the source is 0 terminated the destination will be too.
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.
Currently I'm writing a rather extensive homework assignment that - among other things - reads a file, builds a binary search tree and outputs it.
Somewhere inside all that I've written a recursive method to output the values of the binary search tree in order.
void output(node* n)
{
if(n->leftChild != NULL)
output(n->leftChild);
cout << n->keyAndValue << " || ";
outputString += n->keyAndValue << '|';
if(n->rightChild != NULL)
output(n->rightChild);
}
No problem with that, but you'll notice the line outputString += n->keyAndValue << '|';, because I also want to have all the values inside a char array (I am not allowed to use strings or other more current features of C++) that I can use later on in a different method (e.g. Main method).
The Char-Array is declared as follows:
char *outputString;
This being just one of the ways I've tried. I also tried using the const keyword and just regularly building an array char outputString[]. With the version I've shown you I encounter an error when - later on in the program in a different method - calling the following code:
cout << outputString;
I get the following error:
Unhandled exception at 0x008c2c2a in BST.exe: 0xC00000005: Access Violation reading location 0x5000000000.
Any clue as to how I'd be able to build a dynamic char array, assign values to it numerous times using += and outputting it without triggering an access violation? I am sorry for asking a rather basic question but I am entirely new to C++.
Thanks and Regards,
Dennis
I'm guessing that since you can't use std::string, you also can't use new[].
You can concatenate strings with a function like this:
char *concat(const char *s1, const char *s2)
{
size_t len = strlen(s1) + strlen(s2);
char *result = (char*)malloc(len+1);
strcpy(result, s1);
strcat(result, s2);
return result;
}
This can be done more efficiently, but that probably doesn't matter for homework. And you need to check for errors, etc. etc.
You also need to decide who is going to call free on s1 and s2.
For what it is worth, the efficient version looks like this:
char *concat(const char *s1, const char *s2)
{
size_t len1 = strlen(s1);
size_t len2 = strlen(s2);
char *result = (char*)malloc(len1+len2+1);
memcpy(result, s1, len1);
memcpy(result+len1, s2, len2);
result[len1+len2] = '\0';
return result;
}
It's more efficient because it only walks the input strings once.
+= on pointers does pointer arithmetic, not string concatenation. Eventually you get way beyond your array that outputString was pointing to, and trying to print it leads to a segfault.
Since you can't use std::string, you need to use strcat along with new[] and delete[] and make sure you allocated your original array with new[].
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.
Say I have a vector of null terminates strings some of which may be null pointers. I don't know even if this is legal. It is a learning exercise. Example code
std::vector<char*> c_strings1;
char* p1 = "Stack Over Flow";
c_strings1.push_back(p1);
p1 = NULL; // I am puzzled you can do this and what exactly is stored at this memory location
c_strings1.push_back(p1);
p1 = "Answer";
c_strings1.push_back(p1);
for(std::vector<char*>::size_type i = 0; i < c_strings1.size(); ++i)
{
if( c_strings1[i] != 0 )
{
cout << c_strings1[i] << endl;
}
}
Note that the size of vector is 3 even though I have a NULL at location c_strings1[1]
Question. How can you re-write this code using std::vector<char>
What exactly is stored in the vector when you push a null value?
EDIT
The first part of my question has been thoroughly answered but not the second. Not to my statisfaction at least. I do want to see usage of vector<char>; not some nested variant or std::vector<std::string> Those are familiar. So here is what I tried ( hint: it does not work)
std::vector<char> c_strings2;
string s = "Stack Over Flow";
c_strings2.insert(c_strings2.end(), s.begin(), s.end() );
// char* p = NULL;
s = ""; // this is not really NULL, But would want a NULL here
c_strings2.insert(c_strings2.end(), s.begin(), s.end() );
s = "Answer";
c_strings2.insert(c_strings2.end(), s.begin(), s.end() );
const char *cs = &c_strings2[0];
while (cs <= &c_strings2[2])
{
std::cout << cs << "\n";
cs += std::strlen(cs) + 1;
}
You don't have a vector of strings -- you have a vector of pointer-to-char. NULL is a perfectly valid pointer-to-char which happens to not point to anything, so it is stored in the vector.
Note that the pointers you are actually storing are pointers to char literals. The strings are not copied.
It doesn't make a lot of sense to mix the C++ style vector with the C-style char pointers. Its not illegal to do so, but mixing paradigms like this often results in confused & busted code.
Instead of using a vector<char*> or a vector<char>, why not use a vector<string> ?
EDIT
Based on your edit, it seems like what your'e trying to do is flatten several strings in to a single vector<char>, with a NULL-terminator between each of the flattened strings.
Here's a simple way to accomplish this:
#include <algorithm>
#include <vector>
#include <string>
#include <iterator>
using namespace std;
int main()
{
// create a vector of strings...
typedef vector<string> Strings;
Strings c_strings;
c_strings.push_back("Stack Over Flow");
c_strings.push_back("");
c_strings.push_back("Answer");
/* Flatten the strings in to a vector of char, with
a NULL terminator between each string
So the vector will end up looking like this:
S t a c k _ O v e r _ F l o w \0 \0 A n s w e r \0
***********************************************************/
vector<char> chars;
for( Strings::const_iterator s = c_strings.begin(); s != c_strings.end(); ++s )
{
// append this string to the vector<char>
copy( s->begin(), s->end(), back_inserter(chars) );
// append a null-terminator
chars.push_back('\0');
}
}
So,
char *p1 = "Stack Over Flow";
char *p2 = NULL;
char *p3 = "Answer";
If you notice, the type of all three of those is exactly the same. They are all char *. Because of this, we would expect them all to have the same size in memory as well.
You may think that it doesn't make sense for them to have the same size in memory, because p3 is shorter than p1. What actually happens, is that the compiler, at compile-time, will find all of the strings in the program. In this case, it would find "Stack Over Flow" and "Answer". It will throw those to some constant place in memory, that it knows about. Then, when you attempt to say that p3 = "Answer", the compiler actually transforms that to something like p3 = 0x123456A0.
Therefore, with either version of the push_back call, you are only pushing into the vector a pointer, not the actual string itself.
The vector itself, doesn't know, or care that a NULL char * is an empty string. So in it's counting, it sees that you have pushed three pointers into it, so it reports a size of 3.
I have a funny feeling that what you would really want is to have the vector contain something like "Stack Over Flow Answer" (possibly without space before "Answer").
In this case, you can use a std::vector<char>, you just have to push the whole arrays, not just pointers to them.
This cannot be accomplished with push_back, however vector have an insert method that accept ranges.
/// Maintain the invariant that the vector shall be null terminated
/// p shall be either null or point to a null terminated string
void push_back(std::vector<char>& v, char const* p) {
if (p) {
v.insert(v.end(), p, p + strlen(p));
}
v.push_back('\0');
} // push_back
int main() {
std::vector<char> v;
push_back(v, "Stack Over Flow");
push_back(v, 0);
push_back(v, "Answer");
for (size_t i = 0, max = v.size(); i < max; i += strlen(&v[i]) + 1) {
std::cout << &v[i] << "\n";
}
}
This uses a single contiguous buffer to store multiple null-terminated strings. Passing a null string to push_back results in an empty string being displayed.
What exactly is stored in the vector when you push a null value?
A NULL. You're storing pointers, and NULL is a possible value for a pointer. Why is this unexpected in any way?
Also, use std::string as the value type (i.e. std::vector<std::string>), char* shouldn't be used unless it's needed for C interop. To replicate your code using std::vector<char>, you'd need std::vector<std::vector<char>>.
You have to be careful when storing pointers in STL containers - copying the containers results in shallow copy and things like that.
With regard to your specific question, the vector will store a pointer of type char* regardless of whether or not that pointer points to something. It's entirely possible you would want to store a null-pointer of type char* within that vector for some reason - for example, what if you decide to delete that character string at a later point from the vector? Vectors only support amortized constant time for push_back and pop_back, so there's a good chance if you were deleting a string inside that vector (but not at the end) that you would prefer to just set it null quickly and save some time.
Moving on - I would suggest making a std::vector > if you want a dynamic array of strings which looks like what you're going for.
A std::vector as you mentioned would be useless compared to your original code because your original code stores a dynamic array of strings and a std::vector would only hold one dynamically changable string (as a string is an array of characters essentially).
NULL is just 0. A pointer with value 0 has a meaning. But a char with value 0 has a different meaning. It is used as a delimiter to show the end of a string. Therefore, if you use std::vector<char> and push_back 0, the vector will contain a character with value 0. vector<char> is a vector of characters, while std::vector<char*> is a vector of C-style strings -- very different things.
Update. As the OP wants, I am giving an idea of how to store (in a vector) null terminated strings some of which are nulls.
Option 1: Suppose we have vector<char> c_strings;. Then, we define a function to store a string pi. A lot of complexity is introduced since we need to distinguish between an empty string and a null char*. We select a delimiting character that does not occur in our usage. Suppose this is the '~' character.
char delimiter = '~';
// push each character in pi into c_strings
void push_into_vec(vector<char>& c_strings, char* pi) {
if(pi != 0) {
for(char* p=pi; *p!='\0'; p++)
c_strings.push_back(*p);
// also add a NUL character to denote end-of-string
c_strings.push_back('\0');
}
c_strings.push_back(deimiter);
// Note that a NULL pointer would be stored as a single '~' character
// while an empty string would be stored as '\0~'.
}
// now a method to retrieve each of the stored strings.
vector<char*> get_stored_strings(const vector<char>& c_strings) {
vector<char*> r;
char* end = &c_strings[0] + c_strings.size();
char* current = 0;
bool nullstring = true;
for(char* c = current = &c_strings[0]; c != end+1; c++) {
if(*c == '\0') {
int size = c - current - 1;
char* nc = new char[size+1];
strncpy(nc, current, size);
r.push_back(nc);
nullstring = false;
}
if(*c == delimiter) {
if(nullstring) r.push_back(0);
nullstring = true; // reset nullstring for the next string
current = c+1; // set the next string
}
}
return r;
}
You still need to call delete[] on the memory allocated by new[] above. All this complexity is taken care of by using the string class. I very rarely use char* in C++.
Option 2: You could use vector<boost::optional<char> > . Then the '~' can be replaced by an empty boost::optional, but other other parts are the same as option 1. But the memory usage in this case would be higher.