There's a behavior that I don't understand, the c variable is supposed to increase by one every time an element of the array isn't equal to the one next to it, but there's one increment that's done automatically which increases c by one. I think this is because the last element of the array is compared to something which isn't equal to it. Why is this happening? Or if I'm right what's the thing that the last element is compared to?
#include <iostream>
int main() {
int n, c = 0;
std::cin >> n;
std::string s[n];
for (int i = 0; i < n; i++) {
std::cin >> s[i];
}
for (int i = 0; i < n; i++) {
if (s[i] != s[i+1]) {
c++;
}
}
std::cout << c;
}
std::string s[n]; is creating what is known as a "Variable-Length Array", which is NOT part of standard C++, but is supported as an extension by some compilers. Don't rely on this. Use std::vector instead for a dynamic-length array whose length is not known until runtime.
But, more importantly, your 2nd loop has undefined behavior. The valid indexes for the s[] array's elements are 0..n-1, so on that loop's last iteration, s[i] accesses the last string in the array, and s[i+1] accesses out of bounds, so the value that s[i] is compared to is indeterminate (if the code doesn't just crash outright from accessing invalid memory).
s[s.length()] == ‘\0’
From here: https://www.cplusplus.com/reference/string/string/operator%5B%5D/
“If pos is equal to the string length, the function returns a reference to the null character that follows the last character in the string (which should not be modified).”
Related
I am trying to create an array, which doubles every time it is completely filled.
#include <iostream>
using namespace std;
int* array_doubler(int* array, int size){
int *new_array = new int[size*2];
for(int i = 0; i < size; i++){
new_array[i] = array[i];
}
delete[] array;
return new_array;
}
int main()
{
int N = 10; // Size of array to be created
int *array = new int[0];
for(int i = 0; i < N; i++)
{
if(array[i] == '\0')
array = array_doubler(array, i);
array[i] = i*2;
}
//Printing array elemensts
for(int i = 0; i < N; i++)
cout << array[i] << '\t';
cout << '\n';
return 0;
}
Problem is when I create dynamic memory with new, all the spots have the null character \0 value in them (not just the last spot). i.e. If i write:
int* p = new int[5];
then all the 5 blocks in memory p[0],p[1],p[2],p[3],p[4],p[5] have \0 in them, not just the p[5]. So the if(array[i] == '\0') in my main() calls array_doubler for every single iteration of for loop. I want it to fill the available spots in the array first and when it reaches the last element, then call array_doubler.
Problem is when I create dynamic memory with new, all the spots have the null character \0 value in them (not just the last spot).
Actually they have undefined values in them. 0 is a valid value for them to have, but tomorrow the compiler might suddenly decide that they should all have 1 instead of 0.
If you want to detect the end of an array, then you have to remember how big the array is. C++ doesn't do it for you. Actually, it does do it for you if you use std::vector, but I suppose that's not the point of this exercise.
I'm not sure why you'd want to do this, as std::vector offer this kind of feature, and are more idiomatic of c++ (see isocpp faq on why C-style array are evil).
One of the issue of C-style array is the fact that they don´t know their own size, and that they don't have default value, thus stay uninitialized.
If for some reason you need to not use std::vector, the next best solution would be to wrap the array with it's size in a structure or a class (which is kinda what std::vector is doing), or to initialize your array using std::memset (which is the C function you would use if you were in C).
Do keep in mind that this is not considered as good practices, and that the STL offer plenty of solution when you need containers.
Trying to fill an array using a pointer so each index of array contains its' index. The output I am currently getting is this +DD;D;DD. Could someone explain where I am going wrong on this? Thanks.
#include <iostream>
using namespace std;
int main()
{
int NUMBER_ELEMENTS;
cout << "Enter number of elements: ";
cin >> NUMBER_ELEMENTS;
short array1[NUMBER_ELEMENTS];
short *arrPtr;
arrPtr = array1;
short i = 0;
while(i < NUMBER_ELEMENTS)
{
*arrPtr = i;
arrPtr = arrPtr + 1;
cout << "+" + array1[i];
i++;
}
}
Problem 1: The length of an automatic array can not be given at runtime. It must be known at compile time. In order to create an array with dynamic length, you can use std::vector instead.
explain how this is a variable length array?
You can input different values in different executions of the program. Therefore the length varies. Even the value staying the same is by itself not sufficient. The expression used the array length must be a compile time constant expression.
Problem 2: "+" + array1[i] doesn't do what you probably think it does.
The string literal is an array of characters. array1[i] is an integer. When you add an integer to an array using the plus-operator, the array decays to a pointer to first element of the array, and the the pointer is incremented by number given as the operand.
Therefore "+" + 0 increments the pointer by zero places, so the string printed in the first iteration is "+". "+" + 1 increments the pointer by one places. After the + character, there is only the null terminator, so the printed string is empty. After that iteration, the later iterations overflow the array, and the behaviour of the program is undefined.
I have some code like this :
int n;
cin >> n;
int array[n];
for (int i = 0; i < n; ++i) {
cin >> array[i];
}
int tmp[n - 1];
tmp[0] = 1;
With input : 1 10 I found that the value of array[0] was changed , instead of 10 it has the same with tmp[0].
Then I realized with that input the length of tmp[] became zero. So I print the address of array[] and tmp[] with:
printf("%d\n %d\n", array, tmp);
and found they had the same address.
I want to figure out what will happen if an array has a length of 0; so I tried this:
int array[1];
array[0] = 10;
int tmp[0];
tmp[0] = 1;
address:
array[]: 1363909056
tmp[] : 1363909052
It looks just like the previous code (except the input part). But tmp[0] and array[0] has different values and address now.
And I'm really confused that tmp has smaller address then array.
So my question is:
What will happen if I declare an array of length zero?
Why these two codes works different? (they look the same to me :) )
This is called "undefined behavior". After declaring, in your case
int tmp[0];
The next thing that happens is:
tmp[0]=1;
Since tmp[0] does not exist, this is undefined behavior. Your problem is not that the array was declared with size 0 (that's not exactly kosher on its own merits, but is not the issue here), but the undefined behavior as a result of overwriting memory past the end of the array.
I meet a problem with the char array size. I pass an char array into the function and after run the function, I still want to use sizeof to check the size of the array, it won't give me the new size of the array, but the old size? I would like to know why? Thank you very much!
#include<iostream>
using namespace std;
void replacement(char* arr, int len){
int count=0;
for(int i=0; i<len; i++){
if(arr[i]==' '){
count++;
}
}
int newlen=count*2+len;
//arr[newlen]='\0';
int k=newlen-1;
for(int i=len-1; i>=0; i--){
if(arr[i]!=' '){
arr[k--]=arr[i];
}
else{
arr[k--]='0';
arr[k--]='2';
arr[k--]='%';
}
}
}
int main(){
char arr[]="ab c d e g ";
cout<<sizeof(arr)<<endl;
replacement(arr, sizeof(arr));
int i=0;
while(arr[i]!=NULL) cout<<arr[i];
}
You can't change an array's size. If you want to know the length of the string in the array, use strlen() -- this counts the number of characters before the null terminator.
Even better would be to use C++ std::string class.
Right, so you are trying to replace spaces with "%20", right?
Since C++ (or C) doesn't allow an existing array to be resized, you will either need to have enough space in the first place, or use an array allocated on the heap. Then allocate a new "replacement" string in the replacement function and return that.
The proper C++ method of doing this is of course to use std::string, in which case you could just pass it in as a reference, and do the replacement in the existing variable:
void replacement(std::string* str, int len){
std::string perc20 = "%20";
std::string space = " ";
while((pos = str.find(space, pos)) != std::string::npos)
{
str.replace(pos, space.length(), perc20);
pos += perc20.length();
}
}
Much easier...
You can use sizeof() to find the size of only static arrays when the size is known at compile time. Hence it will always return the size of the array as determined at compile time.
Your program technically has Undefined Behavior because your use of sizeof returns the size in bytes of your char array. But a char implicitly contains a null byte \0. That means the for loop is iterating 1 past the length of the array.
It's recommended that you use std::string along with its size member function instead.
I have to use a dynamic length int array in my program, and want to be able to get the number of objects in it at various points in my code. I am not that familiar with C++, but here is what I have. Why is it not giving me the right length? Thanks.
<#include <iostream>
Using Namespace std;
int length(int*);
void main()
{
int temp[0];
temp[0] = 7;
temp [1] = 10;
temp[2] = '\0';
cout << length(temp) << endl;
}
int length(int* temp)
{
int i = 0;
int count = 0;
while (*temp + i != '\0')
{
count++;
i++;
}
return count;
}
currently it just goes into an endless loop ;_;
In C++ arrays are not dynamic. Your temp array has zero length, and attempting to write to members beyond its length is undefined behaviour. It's most likely not working as it will be writing over some part of the stack.
Either create a fixed size array with enough space to put everything you want to in it, or use a std::vector<int> which is a dynamic data structure.
#include <iostream>
#include <vector>
using namespace std;
int length(int*);
int main () // error: ‘::main’ must return ‘int’
{
int temp[3];
temp[0] = 7;
temp[1] = 10;
// don't use char constants for int values without reason
temp[2] = 0;
cout << length(temp) << endl;
vector<int> vec_temp;
vec_temp.push_back(7);
vec_temp.push_back(10);
cout << vec_temp.size() << endl;
}
int length(int* temp)
{
int i = 0;
int count = 0;
while (*(temp + i) != 0) // *temp + i == (*temp) + i
{
count++;
i++; // don't really need both i and count
}
return count;
}
For the vector, there's no need to specify the size at the start, and you can put a zero in, and finding the length is a simple operation rather than requiring a loop.
Another bug inside your loop was that you were looking at the first member of the array and adding i to that value, rather than incrementing the pointer by i. You don't really need both i and count, so could write that a couple of other ways, either incrementing temp directly:
int length(int* temp)
{
int count = 0;
while (*temp != 0)
{
++count;
++temp;
}
return count;
}
or using count to index temp:
int length(int* temp)
{
int count = 0;
while (temp[count] != 0)
++count;
return count;
}
This approach is a bad idea for a couple of reasons, but first here's some problems:
int temp[0];
This is an array of 0 items, which I don't even think is permitted for stack elements. When declaring an array like this you must specify the maximum number of values you will ever use: E.g. int temp[10];
This is super important! - if you do specify a number less (e.g. [10] and you use [11]) then you will cause a memory overwrite which at best crashes and at worst causes strange bugs that are a nightmare to track down.
The next problem is this line:
while (*temp + i != '\0')
That this line does is take the value stores in the address specified by 'temp' and add i. What you want is to get the value at nth element of the address specified by temp, like so:
while (*(temp + i) != '\0')
So that's what's wrong, but you should take five minutes to think about a better way to do this.
The reasons I mentioned it's a bad idea are:
You need to iterate over the entire array anytime you require its length
You can never store the terminating element (in this case 0) in the array
Instead I would suggest you maintain a separate value that stores the number of elements in the array. A very common way of doing this is to create a class that wraps this concept (a block of elements and the current size).
The C++ standard library comes with a template class named "vector" which can be used for this purpose. It's not quite the same as an array (you must add items first before indexing) but it's very similar. It also provides support for copying/resizing which is handy too.
Here's your program written to use std::vector. Instead of the 'length' function I've added something to print out the values:
#include <vector>
#include <iostream>
void print(std::vector<int> const& vec)
{
using namespace std;
for (size_t i = 0; i < vec.size(); i++)
{
cout << vec[i] << " ";
}
cout << endl;
}
int main()
{
std::vector<int> temp;
temp.push_back(7);
temp.push_back(10);
print(temp);
return 0;
}
You could try:
while (*(temp + i) != '\0')
Your current solution is calculating temp[0] + i (equals 7+i), which apparently is not what you want.
Not only C++ arrays are not dynamic as Pete points out, but only strings (char *) terminate with '\0'. (This is not to say that you can't use a similar convention for other types, but it's rather unusual, and for good reasons: in particular, relying on a terminator symbol requires you to loop through an array to find its size!)
In cases like yours it's better to use the standard library.
#include <vector>
#include <iostream>
int main()
{
std::vector<int> v;
v.push_back(7);
v.push_back(10);
std::cout << v.size() << std::endl;
return 0;
}
If you don't want to use std::vector, try this:
#include <iostream>
using namespace std;
int main () {
int vet[] = {1,2,3,4,5,6};
cout << (sizeof (vet) / sizeof *(vet)) << endl;
return 0;
}
The most common way to get the size of a fixed-length array is something like this:
int temp[256];
int len = sizeof (temp) / sizeof (temp[0]);
// len == 256 * 4 / 4 == 256 on many platforms.
This doesn't work for dynamic arrays because they're actually pointers.
int* temp = new int[256];
int len = sizeof (temp) / sizeof (temp[0]);
// len == 4 / 4 == 1 on many platforms.
For a dynamic-length array if you care about the size, you're best off storing it somewhere when you allocate the array.
The problem with your loop, as pointed out by many is that you have an operator precedence problem here:
*temp + i
should be:
*(temp + i)
But the bigger problem, also pointed out above, is that you don't appear to understand pointers versus fixed-length arrays and are writing off the end of your array.
If you want to use array properly, you have to allocate enough memory for storing values. Once you specified its length, you can't change it. To know array size, you should store it in variable e.g.:
int n;
cin>>n;
int array = new int[n];
int array_length=n;
If you want to change array's length, best way is to use std container, for example std::vector.
Here is the answer to your question
int myarr [] = {1, 2, 3, 4, 5};
int length = sizeof(myarr) / sizeof(myarr[0]);
cout << length;
Because you only allocate space for an array of zero elements.
The following lines
temp [1] = 10;
temp[2] = '\0';
do not allocate more memory or resize the array. You are simply writing data outside the array, corrupting some other part of the application state. Don't do that. ;)
If you want a resizable array, you can use std::vector (and use the push_back member function to insert new values)
A vector also has the size() member function which tells you the current size.
If you want to use the primitive array, you have to track the size yourself. (and, when resizing the array is necessary, copy all elements from the old array to the new, larger one)
To get dynamic behavior in arrays, use a std::vector, or fall back on the old school c style using int * with manual memory allocation (new and delete)[*]
[*] C implementations (discussed in the context of character arrays as C dynamic string length) used malloc, realloc, and free, but these should be avoided in c++ code.
Try this out:
int length(int* temp)
{
int count = 0;
while (*temp != 0 && *temp != -858993460)
{
++count;
++temp;
}
return count;
}