I'm beginner with C and I'm trying to create an array of strings saved like this:
[1][firststring]
[2][secondstring]
[2][thirdstring]
.
.
.
My implementation looks like:
int counter = 0;
char for_pole[50][50];
strcpy(for_pole[counter][50],"hello");
counter++;
//then i want to print it:
printf("my string: %s", for_pole[0][50]); //prints out first string
printf("my string: %s", for_pole[1][50]); //...and second one
but returns segmentation fault.
Should i use some dynamic allocation of memory?
As I've told I'm novice, so sorry for bad question.
I do not see where in your code there is the second string. I see only one string `"hello".
You can write for example the following way
int counter = 0;
char for_pole[50][50];
strcpy(for_pole[counter],"hello");
counter++;
strcpy(for_pole[counter],"world");
//then i want to print it:
printf("my string: %s\n", for_pole[0]); //prints out first string
printf("my string: %s\n", for_pole[1]); //...and second one
As for this expression
for_pole[counter][50]
then 1) index 50 is outside the array because the valid range of indices is 0 - 49 and 2) it has type char instead of to be an array or a pointer to char that is required by function strcpy or by the format specifier %s. of function printf.
When you write
strcpy(for_pole[counter][50],"hello");
You're writing the string "hello" to a char. Moreover, it is on the position 50 of an array that goes from 0 to 49.
you should change it to
strcpy(for_pole[counter],"hello");
because "for_pole[i]" is the i-th string and "for_pole[i][j]" is the j-th character of that string.
Similarly that printf should be
printf("my string: %s", for_pole[0]);
I would add a '\n' to get things a little bit more organized
Finally, the code would be like this
int counter = 0;
char for_pole[50][50];
strcpy(for_pole[counter],"hello");
counter++;
printf("my string: %s\n", for_pole[0]); //prints out first string
printf("my string: %s\n", for_pole[1]); //prints out second string
initialize for_pole variable with zero.
char for_pole[50][50] = {0,};
Jon.e, there are some reasons for SEGFAULT.
Even when there is no bug at all, but rather there are a lack of memory.
Some guys tell you, to pay attention to compiler message and warning, yes, but that's just a part of story.
Related to SEGFAULT there are the following most notable kinds of SEGFAULTS, that I can tell you by my own experience:
1. You are trying to access to indices which do not exist.
Array has the size, exactly same size which you have specified while declaring an array.
Review this code:
/* "Array out of bounds" error
valid indices for array foo
are 0, 1, ... 999 */
int foo[1000];
for (int i = 0; i <= 1000 ; i++)
foo[i] = i;
Here we have one illegal access to an index which does not exist.
2. You are trying to access to the value which does not exist.
In this case memory exists but the value...
/* Illegal memory access if value of n is
not in the range 0, 1, ... 999 */
int n;
int foo[1000];
for (int i = 0; i < n ; i++)
foo[i] = i;
This is also one of the most common pitfalls.
This pitfall is so fundamental, even experienced developers sometime make such mistake. The problem is in uninitialized memory for variable n.
Initialization is process of putting value of the variable its first value, If you want, we can say that it's process of losing virginity of the variable.
There are very chaotic rule of when variable is initialized and when - not. To complex for remembering. Just always initialize variables.
3. You are trying to access to the memory via pointers.
And that pointer does not point anywhere. You are trying to read something from NOWHERE? It does not make sense, does it?
/* Illegal memory access because no memory
is allocated for foo2 */
float *foo, *foo2;
foo = (float*)malloc(1000);
foo2[0] = 1.0;
What happens, if you forget to initialize variable( 2-nd ) or pointer (3-rd case). The answer is broad: from nothing to catastrophe.
Why so? Because memory is the something which has been used by a lot of applications, and it's shared between hundreds of thousands applications running on a computer. So when you create a variable it creates the segment in the memory for that variables, but does not clean that memory (to be precise, in some cases it has been nulled, in others - not) Just assume that is never cleared. So in such situations, no matter the problem caused by variable or by pointer (which is actually is a variable too, but that's another story), we say that, variable stores 'garbage'.
Note, it's not obvious, to find those kind of SEGFAULTS, sometimes it's almost impossible, so you must be ready for such situations, and should use debugger for example gdb, to eliminate the bug.
Also avoid such dummy errors, like. The practice with my students show that it's very very common error.
int n = 100;
int foo[n];
use instead
int foo[100]
or
const int n = 100;
int foo[n];
or
enum N { N = 100 };
int is[N];
Additional references:
Can a const variable be used to declare the size of an array in C?
https://kb.iu.edu/d/aqsj
You misunderstand...
...how strings and array indexing work together. A string is an array of characters. For example this is a string:
char some_string[20] = "Hello";
And this is how you print it:
printf("%s\n", some_string);
And this is how you use strcpy() with it:
strcpy(some_string, "Goodbye");
And this is how you access the i-th character in the string:
/* Not compilable by itself; just an example. */
some_string[i]
Notice how the only time you used [...] was in declaring it and accessing a specific character in the array of characters. However, that is for a single string.
For an array of strings, you declare it as a 2-dimensional array of characters, just as you've done with for_pole:
/* An array of 50 strings, all of which can store a maximum of 49
* characters, plus a null termination character for each string.
*/
char for_pole[50][50];
To use strcpy() with the counter-th string:
strcpy(for_pole[counter], "hello");
And to print the first (0-th) string:
printf("my string: %s", for_pole[0]);
And to print the i-th character of the first string:
printf("string[%d]: %c", i, for_pole[0][i]);
Again, notice how I didn't use [50] every time here. If you have a string, you access it using its name and each character using string[i]. If you have an array of strings, you access the array using its name and each string using array[i] notation. To access the j-th character of the i-th string in the array:
/* Not compilable by itself; just an example. */
array[i][j]
Why your version segfaults
You don't need to use [50] in every expression the way you do. Notice the last expression array[i][j] to access a single character. Since a character is represented by an integer value, the compiler will then convert that integer to a pointer where necessary.
This means that doing something like strcpy(for_pole[counter][50],"hello"); will try to copy the string "hello" to whatever memory address the compiler thinks the character for_pole[counter][50] holds. For example, if the character is the letter A, represented by the integer 65 on my system, then the program would try to copy all 6 bytes of the string "hello" to memory address 65. My program doesn't have permission to write to that memory address, so it segfaults. And that's the problem you are experiencing.
Related
How to print array of string by taking of user input? I am confused between str+i and str[i].
In my program, strings are not printed. It takes 5 strings as input, after that program terminates
#include <stdio.h>
int main()
{
char *str[5];
for (int i = 0; i < 5; i++)
{
scanf("%s", &str[i]);
}
for (int i = 0; i < 5; i++)
{
printf("%s\n", str[i]);
}
}
So firstly, go and read the answers, then come here for complete code. By span of 2 days, I arrived at two solutions to my problem with the help of below answers.
//1.
#include <stdio.h>
#include<stdlib.h>
int main()
{
char* str[5];
for (int i = 0; i < 5; i++)
{
str[i] = (char *)malloc(40 * sizeof(char));
scanf("%s", str[i]);
}
for (int i = 0; i < 5; i++)
{
printf("%s\n", str[i]);
}
}
//2
#include <stdio.h>
#include<stdlib.h>
int main()
{
char str[5][40];
for (int i = 0; i < 5; i++)
{
scanf("%s", &str[i]);
}
for (int i = 0; i < 5; i++)
{
printf("%s\n", str[i]);
}
}
Strings are a painful learning experience in c. Its quite unlike higher level languages.
First off, to answer your explicit question, str[i] is the value of the i-th element in an array. If str points to the array of characters"Hello", then str[1] is the value "e". str + i, on the other hand, is a pointer to the i-th element in an array. In the same situation, where str points to the array of characters "Hello", str + 1 is a pointer to the "e" in that array. str[i] and *(str + i) are identical in every way. In fact, the spec defines a[b] to be (*(a + b)), with all of the behaviors that come with it! (As an aside, C still supports a very ancient notation i[str] which is exactly the same as str[i] because addition of pointers and integers is commutative. You should never use the backwards form, but its worth noting that when I say they are defined to be the same, I mean it!)
Note that I have been very careful to avoid the word "string," and focused on "character arrays" instead. C doesn't technically have a string type. This is important here because you can't do easy things like std::string str[5] (which is a valid C++ notation to create an array of 5 strings) to get a variable length string. You have to make sure you have memory for it. char *str[5] creates an array of 5 char*, but doesn't create any arrays of characters to write to. This is why your code is failing. What is actually happening is each element of str is a pointer to an unspecified address (garbage memory from whatever was there before the variable was created), and scanf is trying to assign into that (nonexistent) array. Bad things happen when you write to somewhere random in memory!
There are two solutions to this. One is to use Serve Lauijssen's approach using malloc. Just please please please please please remember to use free() to deallocate that memory. In nearly any real program, you will not want to leak memory, and using free is a very important habit to get into early on. You should also make sure malloc did not return null. That's another one of those habits. It virtually never returns null on a desktop, but it can. On embedded platforms, it can easily happen. Just check it! (And, from the fact that I have to be reminded of this in the comments suggests I failed to get in the habit early!)
The other approach is to create a multidimensional array of characters. You can use the syntax char str[5][80] to create a 5x80 array of characters.
The exact behavior is a bit of a doozie, but you will find it just happens to behave the way you think it should in your case. You can just use the syntax above, and keep moving. However, you should eventually circle back to understand how this works and what is going on underneath.
C handles the access to these multidimensional arrays in a left to right manner, and it "flattens" the array. In the case of char str[5][80], this will create an array of 400 characters in memory. str[0] will be a char [80] (an 80 character array) which is the first 80 characters in that swath of memory. str[1] will be the next swath of 80 characters, and so on. C will decay an array into a pointer implicitly, so when scanf expects a char*, it will automatically convert the char [80] that is the value of str[i] into a char* that points at the first character of the array. phew
Now, all that explicit "here's what's actually going on" stuff aside, you'll find this does what you want. char str[5][80] will allocate 400 characters of memory, laid out in 5 groups of 80. str[i] will (almost) always turn into a char* pointing at the start of the i-th group of characters. Then scanf has a valid pointer to an array of characters to fill in. Because C "strings" are null-terminated, meaning they end at the first null (character 0 aka '\0') rather than at the end of the memory allocated for it, the extra unused space in the character array simply wont matter.
Again, sorry its so long. This is a source of confusion for basically every C programmer that ever graced the surface of this earth. I am yet to meet a C programmer who was not initially confused by pointers, much less how C handles arrays.
Three other details:
I recommend changing the name from str to strs. It doesn't affect how the code runs, but if you are treating an object as an array, it tends to be more readable if you use plurals. If I was reading code, strs[i] looks like the i-th string in strs, while str[i] looks like the i-th character in a string.
As Bodo points out in comments, using things like scanf("%79s", str[i]) to make sure you don't read too many characters is highly highly highly desirable. Later on, you will be plagued by memory corruptions if you don't ingrain this habit early. The vast majority of exploits you read about in major systems are "buffer overruns" which are where an attacker gets to write too many characters into a buffer, and does something malicious with the extra data as it spills over into whatever happens to be next in the memory space. I'm quite sure you aren't worried about an attacker using your code maliciously at this point in your C career, but it will be a big deal later.
Eventually you will write code where you really do need a char**, that is a pointer to a pointer to a character. The multidimensional array approach won't actually work on that day. When I come across this, I have to create two arrays. The first is char buffer[400] which is my "backing" buffer that holds the characters, and the second is char* strs[5], which holds my strings. Then I have to do strs[0] = buffer + (0 * 80); strs[1] = buffer + (1 * 80); and so on. You don't need this here, but I have needed it in more advanced code.
If you do this, you can also follow the suggestion in the comments to make a static char backing[400]. This creates a block of 400 bytes at compile time which can be used by the function. In general I'd recommend avoiding this, but I include it for completeness. There are some embedded software situations where you will want to use this due to platform limitations. However, this is terribly broken in multithreading situations, which is why many of the standard C functions that relied on static allocated memory now have a variant ending in _r which is re-entrant and threadsafe.
There is also alloca.
Since you have defined char *str[5], so str should be of type char **. So when scanf() expects char *, it is given &str[i] which is of type char **, and so it is not right. That is why your printf("%s\n", str[i]) might not work.
The (str+i) (which is of type char ** and not the same as str[i]) method might work with printf() in the above case because you are reading string values to &str[i]. This could be fine when you have input strings which are short. But then, reading values to &str[i] is not what you intend here, because str[i] is already of type char *. Try giving a very very long string as input (like aaaaaaa...) in the above code, and it will probably give you a Segmentation fault. So technically the method is broken, and you really need to allocate memory to each char * elements in your str array before reading in strings. For example, you can do
for (int i=0; i<5; ++i)
str[i] = (char *) malloc(length_you_wish);
Since each str[i] is of type char * and scanf() expects argument of type char *, use scanf( "%s", str[i] ), omitting & before str[i].
Now, you can printf() with
printf( "%s\n", str[i] ) or printf( "%s\n", *(str+i) ), where str[i] or *(str+i) is of type char *.
Lastly, you need to do
for (int i=0; i<5; ++i)
free(str[i])
Moral: str[i] is a pointer to a specific char array (it holds the address of the first char element of the array), but (str+i) points to str[i] (it refers to the address of str[i], so (str+i) and &str[i] should have same value).
I am trying to test something and I made a small test file to do so. The code is:
void main(){
int i = 0;
char array1 [3];
array1[0] = 'a';
array1[1] = 'b';
array1[2] = 'c';
printf("%s", array1[i+1]);
printf("%d", i);
}
I receive a segmentation error when I compile and try to run. Please let me know what my issue is.
Please let me know what my issue is. ? firstly char array1[3]; is not null terminated as there is no enough space to put '\0' at the end of array1. To avoid this undefined behavior increase the size of array1.
Secondly, array1[i+1] is a single char not string, so use %c instead of %s as
printf("%c", array1[i+1]);
I suggest you get yourself a good book/video series on C. It's not a language that's fun to pick up out of the blue.
Regardless, your problem here is that you haven't formed a correct string. In C, a string is a pointer to the start of a contiguous region of memory that happens to be filled with characters. There is no data whatsoever stored about it's size or any other characteristics. Only where it starts and what it is. Therefore you must provide information as to when the string ends explicitly. This is done by having the very last character in a string be set to the so called null character (in C represented by the escape sequence '\0'.
This implies that any string must be one character longer than the content you want it to hold. You should also never be setting up a string manually like this. Use a library function like strlcpy to do it. It will automatically add in a null character, even if your array is too small (by truncating the string). Alternatively you can statically create a literal string like this:
char array[] = "abc";
It will automatically be null terminated and be of size 4.
Strings need to have a NUL terminator, and you don't have one, nor is there room for one.
The solution is to add one more character:
char array1[4];
// ...
array1[3] = 0;
Also you're asking to print a string but supplying a character instead. You need to supply the whole buffer:
printf("%s", array1);
Then you're fine.
Spend the time to learn about how C strings work, in particular about the requirement for the terminator, as buffer overflow bugs are no joke.
When printf sees a "%s" specifier in the formatting string, it expects a char* as the corresponding argument, but you passed a char value of the array1[i+1] expression. That char got promoted to int but that is still incompatible with char *, And even if it was it has no chance to be a valid pointer to any meaningful character string...
I was doing some pointers and arrays practice in C and I noticed all my four methods returned the same answer.My question is are there disadvantages of using any one of my below methods? I am stunned at how all these four give me the same output. I just noticed you can use a pointer as if it was an array and you can also use an array as if it was a pointer?
char *name = "Madonah";
int i= 0;
for (i=0;i<7; i++){
printf("%c", *(name+i));
}
char name1 [7] = "Madonah";
printf("\n");
int j= 0;
for (j=0;j<7; j++){
printf("%c", name1[j]);
}
char *name2 = "Madonah";
printf("\n");
int k= 0;
for (k=0;k<7; k++){
printf("%c", name2[k]);
}
char name3 [7] = "Madonah";
printf("\n");
int m= 0;
for (m=0;m<7; m++){
printf("%c", *(name+m));
}
Results:
Madonah
Madonah
Madonah
Madonah
It is true that pointers and arrays are equivalent in some context, "equivalent" means neither that they are identical nor even interchangeable. Arrays are not pointers.
It is pointer arithmetic and array indexing that are equivalent, pointers and arrays are different.
which one is preferable and the advantages/Disadvantages?
It depends how you want to use them. If you do not wanna modify string then you can use
char *name = "Madonah";
It is basically equivalent to
char const *name = "Madonah";
*(name + i) and name[i] both are same. I prefer name[i] over *(name + i) as it is neat and used most frequently by C and C++ programmers.
If you like to modify the string then you can go with
char name1[] = "Madonah";
In C, a[b], b[a], *(a+b) are equivalent and there's no difference between these 3. So you only have 2 cases:
char *name = "Madonah"; /* case 1 */
and
char name3 [7] = "Madonah"; /* case 2 */
The former is a pointer which points to a string literal. The latter is an array of 7 characters.
which one is preferred depends on your usage of name3.
If you don't intend to modify then the string then you can use (1) and I would also make it const char* to make it clear and ensure the string literal is not modified accidentally. Modifying string literal is undefined behaviour in C
If you do need to modify it then (2) should be used as it's an array of characters that you can modify. One thing to note is that in case (2), you have explicitly specified the size of the array as 7. That means the character array name3 doesn't have a null terminator (\0) at the end. So it can't be used as a string. I would rather not specify the size of the array and let the compiler calculate it:
char name3 [] = "Madonah"; /* case 2 */
/* an array of 8 characters */
Just in addition to what others said, I will add an image for better illustration. If you have
char a[] = "hello";
char *p = "world";
What happens in first case enough memory is allocated for a (6 characters) on the stack usually, and the string "hello" is copied to memory which starts at a. Hence, you can modify this memory region.
In the second case "world" is allocated somewhere else(usually in read only region), and a pointer to that memory is returned which is simply stored in p. You can't modify the string literal in this case via p.
Here is how it looks:
But for your question stick to notation which is easier, I prefer [].
More info on relationship between arrays and pointers is here.
In all cases, in C pointer + index is the same as pointer[index]. Also, in C an array name used in an expression is treated as a pointer to the first element of the array. Things get a little more mystifying when you consider that addition is commutative, which makes index + pointer and index[pointer] legal also. Compilers will usually generate similar code no matter how you write it.
This allows cool things like "hello"[2] == 'l' and 2["hello"] == 'l'.
it is convenient to use array syntax for random access
int getvalue(int *a, size_t x){return a[x];}
and pointer arithmetic syntax for sequential access
void copy_string(char *dest, const char *src){while((*dest++=*src++));}
Many compilers can optimize sequential accesses to pointers better than using array indices.
For practicing pointer and array idioms in C, those are all valid code blocks and illustrative of the different ways of expressing the same thing.
For production code, you wouldn't use any of those; you would use something both easier for humans to read and more likely to be optimized by the compiler. You'd dispense with the loop entirely and just say:
printf("%s", name);
(Note that this requires name include the \0 character at the end of the string, upon which printf relies. Your name1 and name3 definitions, as written, do not allocate the necessary 8 bytes.)
If you were trying to do something tricky in your code that required one of the more verbose methods you posted, which method you chose would depend on exactly what tricky thing you were trying to do (which you would of course explain in code comments -- otherwise, you would look at your own code six months later and ask yourself, "What the heck was I doing here??"). In general, for plucking characters out of a string, name[i] is a more common idiom than *(name+i).
char name[] = "Madonah";
char* name = "Madonah";
When declared inside a function, the first option yields additional operations every time the function is called. This is because the contents of the string are copied from the RO data section into the stack every time the function is called.
In the second option, the compiler simply sets the variable to point to the address of that string in memory, which is constant throughout the execution of the program.
So unless you're planning to change the contents of the local array (not the original string - that one is read-only), you may opt for the second option.
Please note that all the details above are compiler-implementation dependent, and not dictated by the C-language standard.
can I know which one is easy to use [...]
Try sticking to use the indexing operator [].
One time your code gets more complex and you then notice that things are "more easy" to code using pointer arithmetical expressions.
This typically will be the case when you also start to use the address-of operator: &.
If for example you see your self in the need to code:
char s[] = "Modonah";
char * p = &s[2]; /* This gives you the address of the 'd'. */
you will soon notice that it's "easier" to write:
char * p = s + 2; /* This is the same as char * p = &s[2];. */
An array is just a variable that contains several values, each value has an index, you probably know that already.
Pointers are one of those things you dont need to know about until you realize you need to use them. Pointers are not variables themselves they are literally pointers to variables.
An example of how and why you might want to use a pointer.
You create a variable in one function which you want to use in another function.
You could pass your variable to the new function in the function header.
This effectively COPIES the values from the original variable to a new variable local to the new function. Changes made to it in the new function only change the new variable in the new function.
But what if you wanted changes made in the new function to change the original variable where it is in the first function ?
You use a pointer.
Instead of passing the actual variable to the new function, you pass a pointer to the variable. Now changes made to the variable in the new function are reflected in the original variable in the first function.
This is why in your own example using the pointer to the array and using the actual array while in the same function has identical results. Both of them are saying "change this array".
I'm having a peculiar issue with a simple function I created. This function generates a random number between 0-14, it then creates an array using that randomly generated number as the size and fills it with the char 'x'.
The problem I'm having is that when I call the function it will randomly display symbols or numbers after the x's.
I had originally declared the array as size 15, but figured it was the remaining slots that were causing this display issue. However, it still persists after changing the function.
Here's the current function I'm using:
void application()
{
int randSize, i;
srand((unsigned)time(NULL));
randSize = (rand() % 15);
char array[randSize];
char *bar = array;
for(i=0; i< randSize; i++)
array[i] = 'x';
printf("%s | ", bar);
}
Don't you need to end your strings with \0 in C? ) For example:
char array[randSize + 1];
for (i=0; i < randSize; i++)
array[i] = 'x';
array[i] = '\0';
(updated this because you indeed probably wanted to get a useful part of string of randSize length).
printf("%s | ", bar);
The %s conversion requires a pointer to a 0-terminated character array, but your array is not 0-terminated. Thus the behaviour is undefined, but what usually happens is that printf continues reading and printing characters until it finds a 0 byte or reaches forbidden memory, causing a segmentation fault, whichever happens first.
You're using printf to print a C string. C strings must be null-terminated.
If you want to use array as a string, you'll need to allocate an additional character, and append a \0 to the end:
char array[randSize + 1];
// loop to fill the array
array[i] = '\0';
Since printf expects a null-terminated string, without that it will continue to print past the bounds of your array, reading memory that isn't part of your array.
%s expects a pointer to a null-terminated string. You're not providing that, so you have undefined behaviour and your program is ill-formed.
Read the formatted manual for printf to see how it works and what it expects. In fact, do this for every library function that you are using in your code. No point doing stuff you're not supposed to do and then wondering why it doesn't work; all good libraries will provide specific descriptions of their requirements.
How is this compiling with a variable unless it is declared const? This is an old feature of C99 which is not considered standard. Was not allowed in C prior to that and does not work in C++ either
char array[randSize];
The proper way is
char *array = malloc(randSize);
Am i missing some thing here?
I have the following piece of code in C:
char a[55] = "hello";
size_t length = strlen(a);
char b[length];
strncpy(b,a,length);
size_t length2 = strlen(b);
printf("%d\n", length); // output = 5
printf("%d\n", length2); // output = 8
Why is this the case?
it has to be 'b [length +1]'
strlen does not include the null character in the end of c strings.
You never initialized b to anything. Therefore it's contents are undefined. The call to strlen(b) could read beyond the size of b and cause undefined behavior (such as a crash).
b is not initialized: it contains whatever is in your RAM when the program is run.
For the first string a, the length is 5 as it should be "hello" has 5 characters.
For the second string, b you declare it as a string of 5 characters, but you don't initialise it, so it counts the characters until it finds a byte containing the 0 terminator.
UPDATE: the following line was added after I wrote the original answer.
strncpy(b,a,length);
after this addition, the problem is that you declared b of size length, while it should be length + 1 to provision space for the string terminator.
Others have already pointed out that you need to allocate strlen(a)+1 characters for b to be able to hold the whole string.
They've given you a set of parameters to use for strncpy that will (attempt to) cover up the fact that it's not really suitable for the job at hand (or almost any other, truth be told). What you really want is to just use strcpy instead. Also note, however, that as you've allocated it, b is also a local (auto storage class) variable. It's rarely useful to copy a string into a local variable.
Most of the time, if you're copying a string, you need to copy it to dynamically allocated storage -- otherwise, you might as well use the original and skip doing a copy at all. Copying a string into dynamically allocated storage is sufficiently common that many libraries already include a function (typically named strdup) for the purpose. If you're library doesn't have that, it's fairly easy to write one of your own:
char *dupstr(char const *input) {
char *ret = malloc(strlen(input)+1);
if (ret)
strcpy(ret, input);
return ret;
}
[Edit: I've named this dupstr because strdup (along with anything else starting with str is reserved for the implementation.]
Actually char array is not terminated by '\0' so strlen has no way to know where it sh'd stop calculating lenght of string as as
its syntax is int strlen(char *s)-> it returns no. of chars in string till '\0'(NULL char)
so to avoid this this we have to append NULL char (b[length]='\0')
otherwise strlen count char in string passed till NULL counter is encountered