I am in learning about pointers and strings and how they work. I have a question which answer may I know already. But, in fact, I want to be sure of it.
Let take the following code for example:
char a[200],*p;
gets(a); // cin.get(a,200);
p=a;
strcpy(p,p+1);
printf("%s",a); // c
cout<<a; // cpp
return 0;
Input : working
Output : orking
So what about this? Why do the array modified without actually modify itself?
char a[200],*p;
gets(a); // cin.get(a,200);
p=a;
p++;
printf("%s",a); // cout<<a;
return 0;
Input : working
Output : working
In this case, why the array did not modify? What has the function strcpy special? And which other functions has this propriety?
Throughout the text, I have referred to the functions of the strings set in the program.
Snippet 1
char a[200],*p;
Defined a block of 200 characters ,a, and a pointer, p, to a block of one or more characters.
gets(a); // cin.get(a,200);
Read some user input into memory block a
p=a;
Points p at block a. p can now be used as a reference to the same thing as a
strcpy(p,p+1);
Copy part of the memory block over itself. NOTE: this invokes undefined behaviour because the to and from buffers in strcpy overlap (See http://en.cppreference.com/w/cpp/string/byte/strcpy). The results may be unusual, ranging from "Looks like it works!" to more fantastic, such as the computer making it rain unicorns.
So, say a is created at memory location 10. Data is read into the block starting at location 10. Then p is set to point at location ten. Then the memory starting at location 11 is copied over the memory starting at location 10. Since a is at 10, printing a prints the altered data. Since p is still pointing at 10, printing p will print the same thing as a.
Snippet 2:
char a[200],*p;
Defined a block of 200 characters ,a, and a pointer, p, to a block of one or more characters.
gets(a); // cin.get(a,200);
Read some user input into memory block a
p=a;
Points p at block a. p can now be used as a reference to the same thing as a
p++;
Moved where p pointed over one slot to the right. Copied nothing. Just changed the address that p points at.
So, say a is created at memory location 10. Data is read into the block starting at location 10. Then p is set to point at location 10. Then p is set to point at memory location 11. The values in memory block a are unchanged. Since a is still at 10, printing a prints the unchanged data. Since p is now pointing at 11, printing p will print the from the second character of a onward.
You are not modifying anything in the second code. Why should it be modified?
Piano points to the first element of a. Then you increment it. So please points to the second element. Then you print a, who wasn't modified.
As for the first one, from strcpy reference :
To avoid overflows, the size of the array pointed by destination shall be long enough to contain the same C string as source(including the terminating null character), and should not overlap in memory with source
I think you have what is called an overlap.
You copy a source to a destination that's overlapping in memory.
Related
If I take a block of memory by the following line .
int* a = new int[10];
Then for freeing the memory , the code would be
delete [] a;
But if I take a pointer of single memory segment like the following
int* a = new int;
And then insert a array of data like following .
for(int i=0;i<10;i++)
{
a[i]= i ;
}
So to free the first memory segment that pointer "a" is pointing, the code would be like following
delete a;
But Here I inserted 9 more data from the memory that pointer "a" is pointing .So I am using actually 10 memory segment here . how can I free all this 10 memory ? Please help me to get the answer .
how can I free all this 10 memory ?
You can't and you shouldn't because the moment you tried to "insert a array of data like following" you have entered Undefined Behavior land for writing to a location that you didn't allocate with new in the first place. You asked for a single int, you got a single int. Don't write past it.
You allocated only one int. Memory from a[1] to a[9] may be assigned to other objects and you may corrupt them.
P.S. Btw you can not free memory that you did not allocate in any case.
new int allocates space for one int value. You cannot legally pretend that it's an array of ten int values, and the compiler won't generate code to expand the allocated memory if you go out of bounds. The code you wrote produces undefined behavior. The fact that it compiled and ran doesn't change that; sooner or later it will cause problems.
To allocate an array that can hold 10 int values, use your first expression: new int[10]. That will allocate space for ten int values.
To allocate an array that can be expanded at will, use std::vector<int>.
Ok a very basic question, but I am stuck at it. I cannot figure out the extra mysterious value at the end of the array. I just tried to traverse the array through its base address, plain and simple. But the garbage value at the end of the array remains constant every time I execute it
g++ complier 64 bit machine.
#include<iostream>
#include<bits/stdc++.h>
using namespace std;
void printarray(int * arr){
while(*(arr)){ cout<<*arr<<" "; arr++; }
cout<<endl;
}
int main(){
int arr[] = { 3,2,4,1,5 };
printarray(arr); // prints 3,2,4,1,5,32765
return 0;
}
EDIT 1: I understand that the while will terminate whenever it comes across a 0 in the array , or if no 0 is found will go insane . But still I would want you guys to look at the following test cases
3,2,0,4,1,5 // outputs 3,2
3,2,4,1,5,7,8,9 // outputs same array correctly
3,2,4,1,5,7,8,9,10 // outputs array+ 1 varying garbage at last
//observations, this method works for even sized arrays not containing
//0, while for odd it emits an additional garbage value.
My question is if the while only breaks at 0 , why does it break at
arrayLength+1 everytime ? Also what's with this even odd length ?
An array is not terminated in any special way. What you are trying to do is not working because while(*arr) relies on a wrong assumption: that there is a 0 element at the end of the array (a sentinel) but you declare the array a {3,2,4,1,5} so there is no end element.
I think your misconcept comes from the fact that you are not getting the point that an int* is just a memory address, whenever you increment it by ++arr you are basically adjusting it by sizeof(int). Nothing more, when you reach the end of the array then the address just points after the array, to whatever value could be there.
You get the extra element(s) because there is nothing about a pointer that tells the compiler how many elements the pointer points to. When you do
while(*(arr)){...}
The while loop will continue running untill *arr == 0. Since you array doesn't contain a 0 it will keep going past the end of the array untill it finds a 0. This is undefined behavior as you are accessing memory you do not own with that pointer.
I think you may be confusing how char arrays(c-strings) work compared to other data types. When you have
char arr[] = "something";
while(*(arr)){...}
This ends at the end of the array as c-strings get a null terminator(0) added to the end of the string automatically. This allows you to do things like the above loop as you know that null terminator will be there and if it is not then that is on the person you created the string.
An array decays into a pointer when passed to a function, and this function knows nothing about the array's length.
That while(*arr) stuff is incorrect. It will stop only when some value this pointer points to is zero. But who said a zero is placed at the end of an array?? When you increment arr, you can easily get out of bounds of your array (the function doesn't know its size!), and then *arr will give you whatever the heck is stored at the memory address arr points to at the moment.
To iterate over an array, pass the array itself and its length. Otherwise this will iterate over and over until the value of *arr will be zero.
Pointers aren't terminated in C++ by any character.
This works on other types like char* only because it's terminated by an \0(0).
An INT-Array u need to count the Elements before you can pass them into something like that, for example here with the ending 5 from your Pointer:
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
void printarray(int * arr){
int *saved=arr;
while(*saved!=5)
{
cout<<*saved++<<" ";
}
*saved=0;
cout<<endl;
}
int main(){
int arr[] = { 3, 2, 4, 1, 5 };
printarray(arr); // prints now without ending 5.
return 0;
}
Otherwise, you need to pass the counter of elements.
There is no way C++ could know how many elements your pointer points to.
int main() {
int* i = new int(1);
i++;
*i=1;
delete i;
}
Here is my logic:
I increment I by 1, and then assign a value to it. Then I delete the I, so I free the memory location while leaking the original memory. Where is my problem?
I also tried different versions. Every time, as long as I do the arithmetics and delete the pointer, my program crashes.
What your program shows is several cases of undefined behaviour:
You write to memory that hasn't been allocated (*i = 1)
You free something that you didn't allocate, effectively delete i + 1.
You MUST call delete on exactly the same pointer-value that you got back from new - nothing else. Assuming the rest of your code was valid, it would be fine to do int *j = i; after int *i = new int(1);, and then delete j;. [For example int *i = new int[2]; would then make your i++; *i=1; valid code]
Who allocates is who deallocates. So you should not be able to delete something you did not new by yourself. Furthermore, i++;*i=1; is UB since you may access a restricted memory area or read-only memory...
The code made no sense . I think You have XY problem. If you could post your original problem there will be more chance to help you.
In this case you need to have a short understanding how the heap memory management works. in particular implementation of it, when you allocate an object you receive a pointer to the start of the memory available to you to work with. However, the 'really' allocated memory starts a bit 'earlier'. This means the allocated block is a bit more than you have requested to allocate. The start of the block is the address you have received minus some offset. Thus, when you pass the incremented pointer to the delete it tries to find the internal information at the left side of it. And because your address is now incremented this search fails what results in a crash. That's in short.
The problem lies here:
i++;
This line doesn't increment the value i points to, but the pointer itself by the number of bytes an int has (4 on 32-bit platform).
You meant to do this:
(*i)++;
Let's take it step by step:
int* i = new int(1); // 1. Allocate a memory.
i++; // 2. Increment a pointer. The pointer now points to
// another location.
*i=1; // 3. Dereference a pointer which points to unknown
// memory. This could cause segmentation fault.
delete i; // 4. Delete the unknown memory which is undefined
// behavior.
In short: If you don't own a piece of memory you can't do arithmetic with it neither delete it!
consider this code:
double *pi;
double j;
pi = &j;
pi[3] = 5;
I don't understand how is that possible that I can perform the last line here.
I set pi to the reference of j, which is a double variable, and not a double [] variable. so how is this possible that I can perform an array commands on it?
consider this code:
char *c = "abcdefg";
std::cout << &(c[3]) << endl;
the output is "defg". I expected that I will get a reference output because I used &, but instead I got the value of the char * from the cell position to the end. why is that?
You have two separate questions here.
A pointer is sometimes used to point to an array or buffer in memory. Therefore it supports the [] syntax. In this case, using pi[x] where x is not 0 is invalid as you are not pointing to an array or buffer.
Streams have an overload for char pointers to treat them as a C-style string, and not output their address. That is what is happening in your second case. Try std::cout << static_cast<const void *>(&(c[3])) << endl;
Pointers and arrays go hand in hand in C (sort of...)
pi[3] is the same as *(pi + 3). In your code however this leads to Undefined Behavior as you create a pointer outside an object bounds.
Also be careful as * and & are different operators depending on in which kind of expression the appear.
That is undefined behavior. C++ allows you to do things you ought not to.
There are special rules for char*, because it is often used as the beginning of a string. If pass a char* to cout, it will print whatever that points to as characters, and stop when it reaches a '\0'.
Ok, so a few main things here:
A pointer is what it is, it points to a location in the memory. So therefore, a pointer can be an array if you whish.
If you are working with pointers (dangerous at times), this complicates things. You are writing on p, which is a pointer to a memory location. So, even though you have not allocated the memory, you can access the memory as an array and write it. But this gives us the question you are asking. How can this be? well, the simple answer is that you are accessing a zone of memory where the variable you have created has absolutely no control, so you could possibly be stepping on another variable (if you have others) or simply just writting on memory that has not been used yet.
I dont't understand what you are asking in the second question, maybe you could explain a little more? Thanks.
The last line of this code...
double *pi;
double j;
pi = &j;
pi[3] = 5;
... is the syntactic equivalent to (pi + 3) = 5. There is no difference in how a compiler views a double[] variable and a double variable.
Although the above code will compile, it will cause a memory error. Here is safe code that illustrates the same concepts...
double *pi = new double[5]; // allocate 5 places of int in heap
double j;
pi[3] = 5; // give 4th place a value of 5
delete pi; // erase allocated memory
pi = &j; // now get pi to point to a different memory location
I don't understand how is that possible that I can perform the last
line here. I set pi to the reference of j
Actually, you're setting your pointer pi, to point to the memory address of j.
When you do pi[3], you're using a non-array variable as an array. While valid c++, it is inherently dangerous. You run the risk of overwriting the memory of other variables, or even access memory outside your process, which will result in the operating system killing your program.
When that's said, pi[3] means you're saying "give me the slot third down from the memory location of pi". So you're not touching pi itself, but an offset.
If you want to use arrays, declare them as such:
double pi[5]; //This means 5 doubles arrayed aside each other, hence the term "array".
Appropos arrays, in c++ it's usually better to not use raw arrays, instead use vectors(there are other types of containers):
vector<double> container;
container.push(5.25); //"push" means you add a variable to the vector.
Unlike raw arrays, a container such as a vector, will keep it's size internally, so if you've put 5 doubles in it, you can call container.size(), which will return 5. Useful in for loops and the like.
About your second question, you're effectively returning a reference to a substring of your "abcdefg" string.
&([3]) means "give me a string, starting from the d". Since c-style strings(which is what char* is called) add an extra NULL at the end, any piece of code that takes these as arguments(such as cout) will keep reading memory until they stumble upon the NULL(aka a 0). The NULL terminates the string, meaning it marks the end of the data.
Appropos, c-style strings are the only datatype that behaves like an array, without actually being one. This also means they are dangerous. Personally I've never had any need to use one. I recommend using modern strings instead. These newer, c++ specific variables are both safe to use, as well as easier to use. Like vectors, they are containers, they keep track of their size, and they resize automatically. Observe:
string test = "abcdefg";
cout<<test.size()<<endl;//prints 7, the number of characters in the array.
test.append("hijklmno");//appends the string, AND updates the size, so subsequent calls will now return 15.
#include<iostream.h>
#include<conio.h>
int *p;
void Set()
{
int i;
i=7;
p=&i;
}
int Use()
{
double d;
d=3.0;
d+=*p;
//if i replace the above 3 statements with --> double d=3.0+*p; then output is 10 otherwise the output is some random value(address of random memory location)
//why is this happening? Isn't it the same?
return d;
}
void main()
{
clrscr();
Set();
cout<<Use();
getch();
}
My question is as mentioned in comments above.I want to know the exact reason for the difference in outputs.In the above code output is some address of random memory location and i understand that is because of i is a local variable of Set() but then how is it visible in the second case that is by replacing it with double d=3.0+*p; because then the output comes 10( 7+3 ) although 7 should not have been visible?
The result of using pointer p is undefined, it could also give you a segfault or just return 42. The technical reason behind the results your'e getting are probably:
i inside Set is placed on the stack. The value 7 ist stored there and p points to that location in memory. When you return from Set value remains in memory: the stack is not "destroyed", it's just the stack pointer which is reset. p still points to this location which still contains the integer representation of "3".
Inside Use the same location on the stack is reused for d.
When the compiler is not optimizing, in the first case (i.e. the whole computation in one line), it first uses the value 7 (which is still there in memory with p pointing to it), does the computation, overwrites the value (since you assign it to d which is at the same location) and returns it.
In the second case, it first overwrites the value withe the double value 3.0 and then takes the first 4 bytes interpreted as integer value for evaluating *p in d+=*p.
This case shows why returning pointers/references to local variables is such a bad thing: when writing the function Set you could even write some kind of unit tests and they would not detect the error. It might get unnoticed just until the software goes into production and has to perform some really critical task and just fails then.
This applies to all kindes of "undefined behaviour", especially in "low level" languages like C/C++. The bad thing is that "undefined" may very well mean "perfectly working until it's too late"...
After exiting function Set the value of pointer p becomes invalid due to destroying local variable i. The program has undefined behavior.