I was wondering if and how one would compare an array of char pointers to a string.
So say I have this array of char pointers:
char *input[20];
And each index of input contained a string, for example, input[0] contained hello. What would I use if I needed to do a comparison to find a keyword contained within the input array?
Not really clear what your problem is here. But something like:
for ( int i = 0; i < 20; i++ ) {
if ( strcmp( input[i], "keyword" ) == 0 ) {
// found - do something
}
}
But in C++ you would be better off using std::string and std::vector rather than messing around with C-style arrays and pointers.
Related
I have a char array and I want to find out the number of contents in it.
For example, my array is:
char myArray[10];
And after input it's content is:
ABC
Now I want to store in a variable 'size', the size of the area related to content. So, in this case:
size = 3
How do I find that?
A naive way of doing this would be to look for the null-terminating character \0, this is already implemented for you in the C-function strlen, so there are two ways of doing this:
int StringLength( const char* str, int maxLength )
{
for( int i = 0; i < maxLength; ++i )
{
if( str[i] == '\0' )
return i;
}
return -1;
}
Or you could just call strlen as follows:
int iLength = strlen( myArray );
However, as you have tagged this c++, the best way to do this would be to not deal with C-style character arrays and instead use the extremely useful std::string class.
strlen(myArray) is what you want.
Try this:
int len = strlen(myArray).
strlen is a part of stdlib.h library. Don't forget to declare it in your program.
defining array as char myArray[10]; will not always initialize it's content to zeros, so depending on how you fill it with ABC you either can or cannot find the corect lenght. In worst case regular strlen() will always report numbers >10, or even result in read access vialation. I try initialize it like char myArray[10] = {}; first
This question already has answers here:
Why can't I write to a string literal while I *can* write to a string object?
(4 answers)
Closed 8 years ago.
Someone can tell me different between this code:
char *s1 ="Con Chim Non";
and this one:
char *s=new char[100];
gets(s);
Then, I add the word: "Con Chim Non".
After, I build a code that change value of the pointer. In first code, I got a problem about the address. The second is right. And this is my code:
void Strlwr(char *s)
{
if (s == NULL )
return ;
for (int i = 0; s[i] != '\0'; ++i)
{
if ( s[i] <= 'Z' && s[i] >= 'A')
s[i] = s[i]+32;
}
}
And someone can tell me why the first is wrong.
The first example:
char *s1 ="Con Chim Non";
You declare a pointer to a text literal, which is constant. Text literals cannot be modified.
The proper syntax is:
char const * s1 = "Con Chim Non";
Note the const.
In your second example, you are declaring, reserving memory for 100 characters in dynamic memory:
char *s=new char[100];
gets(s);
You are then getting an unknown amount of characters from the input and placing them into the array.
Since you are programming in the C++ language, you should refrain from this kind of text handling and use the safer std::string data type.
For example, the gets function will read an unknown amount of characters from the console into an array. If you declare an array of 4 characters and type in 10, you will have a buffer overflow, which is very bad.
The std::string class will expand as necessary to contain the content. It will also manage memory reallocation.
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.
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.
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.