I have a simple error that I know lies underneath my C code's memory usage, because the iterator in a simple for loop changes drastically after receiving user input on the command line:
int i = 1;
char input[] = "";
for (i = 1; i <= 5; i++) {
printf("i %d\n> ", i);
scanf("%s", input);
printf("input %s\ni %d\n", input, i);
}
The output should be simple enough:
i 1
> <receive input>
input <input>
i 1
to be repeated 5 times.
However, the iterator 'i' changes to anything but what is expected when any input is received.
An example output:
i 1
> 45
input 45
i 53
I have a hunch that this comes from memory access in the compiler. Anything helps!
scanf("%s", input);
From the docs of scanf:
%s Matches a sequence of bytes that are not white-space characters. The application shall ensure that the corresponding argument is a pointer to the initial byte of an array of char, signed char, or unsigned char large enough to accept the sequence and a terminating null character code, which shall be added automatically.
You are the application. char input[] = ""; is only 1 byte big.
Any character returned by the scanf will result in overflowing input due to the null terminator. And will write over the next variable in memory.
Try:
char input[100] = "";
scanf("%100s", input);
Look at how your local variables are declared:
int i = 1;
char input[] = "";
input is a zero-length string, and there's no room allocated for the input you're about to ask for. When you do:
scanf("%s", input);
the input gets written into the array pointed to by input, but since there was no space reserved for that, whatever happens to be after the array that input refers to gets written over. In this case, that's I.
To solve the problem, you need to make sure that there's enough room for the input at the location where you're putting it. Also, you should limit the allowable length of the input so that the user can't enter more data than the size of the space you've reserved.
Related
For an instance if I store ABCDE from scanf function, the later printf function gives me ABCDE as output. So what is the point of assigning the size of the string(Here 4).
#include <stdio.h>
int main() {
int c[4];
printf("Enter your name:");
scanf("%s",c);
printf("Your Name is:%s",c);
return 0;
}
I'll start with, don't use int array to store strings!
int c[4] allocates an array of 4 integers. An int is typically 4 bytes, so usually this would be 16 bytes (but might be 8 or 32 or something else on some platforms).
Then, you use this allocation first to read characters with scanf. If you enter ABCDE, it uses up 6 characters (there is an extra 0 byte at the end of the string marking the end, which needs space too), which happens to fit into the memory reserved for array of 4 integers. Now you could be really unlucky and have a platform where int has a so called "trap representation", which would cause your program to crash. But, if you are not writing the code for some very exotic device, there won't be. Now it just so happens, that this code is going to work, for the same reason memcpy is going to work: char type is special in C, and allows copying bytes to and from different types.
Same special treatment happens, when you print the int[4] array with printf using %s format. It works, because char is special.
This also demonstrates how very unsafe scanf and printf are. They happily accept c you give them, and assume it is a char array with valid size and data.
But, don't do this. If you want to store a string, use char array. Correct code for this would be:
#include <stdio.h>
int main() {
char c[16]; // fits 15 characters plus terminating 0
printf("Enter your name:");
int items = scanf("%15s",c); // note: added maximum characters
// scanf returns number of items read successfully, *always* check that!
if (items != 1) {
return 1; // exit with error, maybe add printing error message
}
printf("Your Name is: %s\n",c); // note added newline, just as an example
return 0;
}
The size of an array must be defined while declaring a C String variable because it is used to calculate how many characters are going to be stored inside the string variable and thus how much memory will be reserved for your string. If you exceed that amount the result is undefined behavior.
You have used int c , not char c . In C, a char is only 1 byte long, while a int is 4 bytes. That's why you didn't face any issues.
(Simplifying a fair amount)
When you initialize that array of length 4, C goes and finds a free spot in memory that has enough consecutive space to store 4 integers. But if you try to set c[4] to something, C will write that thing in the memory just after your array. Who knows what’s there? That might not be free, so you might be overwriting something important (generally bad). Also, if you do some stuff, and then come back, something else might’ve used that memory slot (properly) and overwritten your data, replacing it with bizarre, unrelated, and useless (to you) data.
In C language the last of the string is '\0'.
If you print with the below function, you can see the last character of the string.
scanf("%s", c); add the last character, '\0'.
So, if you use another function, getc, getch .., you should consider adding the laster character by yourself.
#include<stdio.h>
#include<string.h>
int main(){
char c[4+1]; // You should add +1 for the '\0' character.
char *p;
int len;
printf("Enter your name:");
scanf("%s", c);
len = strlen(c);
printf("Your Name is:%s (%d)\n", c, len);
p = c;
do {
printf("%x\n", *(p++));
} while((len--)+1);
return 0;
}
Enter your name:1234
Your Name is:1234 (4)
31
32
33
34
0 --> last character added by scanf("%s);
ffffffae --> garbage
This is my code for two functions in C:
// Begin
void readTrain(Train_t *train){
printf("Name des Zugs:");
char name[STR];
getlinee(name, STR);
strcpy(train->name, name);
printf("Name des Drivers:");
char namedriver[STR];
getlinee(namedriver, STR);
strcpy(train->driver, namedriver);
}
void getlinee(char *str, long num){
char c;
int i = 0;
while(((c=getchar())!='\n') && (i<num)){
*str = c;
str++;
i++;
}
printf("i is %d\n", i);
*str = '\0';
fflush(stdin);
}
// End
So, with void getlinee(char *str, long num) function I want to get user input to first string char name[STR] and to second char namedriver[STR]. Maximal string size is STR (30 charachters) and if I have at the input more than 30 characters for first string ("Name des Zuges"), which will be stored in name[STR], after that I input second string, which will be stored in namedriver, and then printing FIRST string, I do not get the string from the user input (first 30 characters from input), but also the second string "attached" to this, I simply do not know why...otherwise it works good, if the limit of 30 characters is respected for the first string.
Here my output, when the input is larger than 30 characters for first string, problem is in the row 5 "Zugname", why I also have second string when I m printing just first one...:
Name des Zugs:aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
i is 30
Name des Drivers:xxxxxxxx
i is 8
Zugname: aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaxxxxxxxx
Drivername: xxxxxxxx
I think your issue is that your train->name is not properly terminated with '\0', as a consequence when you call printf("%s", train->name) the function keeps reading memory until it finds '\0'. In your case I guess your structure looks like:
struct Train_t {
//...
char name[STR];
char driver[STR];
//...
};
In getlinee() function, you write '\0' after the last character. In particular, if the input is more than 30 characters long, you copy the first 30 characters, then add '\0' at the 31-th character (name[30]). This is a first buffer overflow.
So where is this '\0' actually written? well, at name[30], even though your not supposed to write there. Then, if you have the structure above when you do strcpy(train->name, name); you will actually copy a 31-bytes long string: 30 chars into train->name, and the '\0' will overflow into train->driver[0]. This is the second buffer overflow.
After this, you override the train->driver buffer so the '\0' disappears and your data in memory basically looks like:
train->name = "aaa...aaa" // no '\0' at the end so printf won't stop reading here
train->driver = "xxx\0" // but there
You have an off-by-one error on your array sizes -- you have arrays of STR chars, and you read up to STR characters into them, but then you store a NUL terminator, requiring (up to) STR + 1 bytes total. So whenever you have a max size input, you run off the end of your array(s) and get undefined behavior.
Pass STR - 1 as the second argument to getlinee for the easiest fix.
Key issues
Size test in wrong order and off-by-one. ((c=getchar())!='\n') && (i<num) --> (i+1<num) && ((c=getchar())!='\n'). Else no room for the null character. Bad form to consume an excess character here.
getlinee() should be declared before first use. Tip: Enable all compiler warnings to save time.
Other
Use int c; not char c; to well distinguish the typical 257 different possible results from getchar().
fflush(stdin); is undefined behavior. Better code would consume excess characters in a line with other code.
void getlinee(char *str, long num) better with size_t num. size_t is the right size type for array sizing and indexing.
int i should be the same type as num.
Better code would also test for EOF.
while((i<num) && ((c=getchar())!='\n') && (c != EOF)){
A better design would return something from getlinee() to indicate success and identify troubles like end-of-file with nothing read, input error, too long a line and parameter trouble like str == NULL, num <= 0.
I believe you have a struct similar to this:
typedef struct train_s
{
//...
char name[STR];
char driver[STR];
//...
} Train_t;
When you attempt to write a '\0' to a string that is longer than STR (30 in this case), you actually write a '\0' to name[STR], which you don't have, since the last element of name with length STR has an index of STR-1 (29 in this case), so you are trying to write a '\0' outside your array.
And, since two strings in this struct are stored one after another, you are writing a '\0' to driver[0], which you immediately overwrite, hence when printing out name, printf doesn't find a '\0' until it reaches the end of driver, so it prints both.
Fixing this should be easy.
Just change:
while(((c=getchar())!='\n') && (i<num))
to:
while(((c=getchar())!='\n') && (i<num - 1))
Or, as I would do it, add 1 to array size:
char name[STR + 1];
char driver[STR + 1];
I'm new to C and I'm trying to write a program that prints the ASCII value for every letter in a name that the user enters. I attempted to store the letters in an array and try to print each ASCII value and letter of the name separately but, for some reason, it only prints the value of the first letter.
For example, if I write "Anna" it just prints 65 and not the values for the other letters in the name. I think it has something to do with my sizeof(name)/sizeof(char) part of the for loop, because when I print it separately, it only prints out 1.
I can't figure out how to fix it:
#include <stdio.h>
int main(){
int e;
char name[] = "";
printf("Enter a name : \n");
scanf("%c",&name);
for(int i = 0; i < (sizeof(name)/sizeof(char)); i++){
e = name[i];
printf("The ASCII value of the letter %c is : %d \n",name[i],e);
}
int n = (sizeof(name)/sizeof(char));
printf("%d", n);
}
Here's a corrected, annotated version:
#include <stdio.h>
#include <string.h>
int main() {
int e;
char name[100] = ""; // Allow for up to 100 characters
printf("Enter a name : \n");
// scanf("%c", &name); // %c reads a single character
scanf("%99s", name); // Use %s to read a string! %99s to limit input size!
// for (int i = 0; i < (sizeof(name) / sizeof(char)); i++) { // sizeof(name) / sizeof(char) is a fixed value!
size_t len = strlen(name); // Use this library function to get string length
for (size_t i = 0; i < len; i++) { // Saves calculating each time!
e = name[i];
printf("The ASCII value of the letter %c is : %d \n", name[i], e);
}
printf("\n Name length = %zu\n", strlen(name)); // Given length!
int n = (sizeof(name) / sizeof(char)); // As noted above, this will be ...
printf("%d", n); // ... a fixed value (100, as it stands).
return 0; // ALWAYS return an integer from main!
}
But also read the comments given in your question!
This is a rather long answer, feel free to skip to the end for the code example.
First of all, by initialising a char array with unspecified length, you are making that array have length 1 (it only contains the empty string). The key issue here is that arrays in C are fixed size, so name will not grow larger.
Second, the format specifier %c causes scanf to only ever read one byte. This means that even if you had made a larger array, you would only be reading one byte to it anyway.
The parameter you're giving to scanf is erroneous, but accidentally works - you're passing a pointer to an array when it expects a pointer to char. It works because the pointer to the array points at the first element of the array. Luckily this is an easy fix, an array of a type can be passed to a function expecting a pointer to that type - it is said to "decay" to a pointer. So you could just pass name instead.
As a result of these two actions, you now have a situation where name is of length 1, and you have read exactly one byte into it. The next issue is sizeof(name)/sizeof(char) - this will always equal 1 in your program. sizeof char is defined to always equal 1, so using it as a divisor causes no effect, and we already know sizeof name is equal to 1. This means your for loop will only ever read one byte from the array. For the exact same reason n is equal to 1. This is not erroneous per se, it's just probably not what you expected.
The solution to this can be done in a couple of ways, but I'll show one. First of all, you don't want to initialize name as you do, because it always creates an array of size 1. Instead you want to manually specify a larger size for the array, for instance 100 bytes (of which the last one will be dedicated to the terminating null byte).
char name[100];
/* You might want to zero out the array too by eg. using memset. It's not
necessary in this case, but arrays are allowed to contain anything unless
and until you replace their contents.
Parameters are target, byte to fill it with, and amount of bytes to fill */
memset(name, 0, sizeof(name));
Second, you don't necessarily want to use scanf at all if you're reading just a byte string from standard input instead of a more complex formatted string. You could eg. use fgets to read an entire line from standard input, though that also includes the newline character, which we'll have to strip.
/* The parameters are target to write to, bytes to write, and file to read from.
fgets writes a null terminator automatically after the string, so we will
read at most sizeof(name) - 1 bytes.
*/
fgets(name, sizeof(name), stdin);
Now you've read the name to memory. But the size of name the array hasn't changed, so if you used the rest of the code as is you would get a lot of messages saying The ASCII value of the letter is : 0. To get the meaningful length of the string, we'll use strlen.
NOTE: strlen is generally unsafe to use on arbitrary strings that might not be properly null-terminated as it will keep reading until it finds a zero byte, but we only get a portable bounds-checked version strnlen_s in C11. In this case we also know that the string is null-terminated, because fgets deals with that.
/* size_t is a large, unsigned integer type big enough to contain the
theoretical maximum size of an object, so size functions often return
size_t.
strlen counts the amount of bytes before the first null (0) byte */
size_t n = strlen(name);
Now that we have the length of the string, we can check if the last byte is the newline character, and remove it if so.
/* Assuming every line ends with a newline, we can simply zero out the last
byte if it's '\n' */
if (name[n - 1] == '\n') {
name[n - 1] = '\0';
/* The string is now 1 byte shorter, because we removed the newline.
We don't need to calculate strlen again, we can just do it manually. */
--n;
}
The loop looks quite similar, as it was mostly fine to begin with. Mostly, we want to avoid issues that can arise from comparing a signed int and an unsigned size_t, so we'll also make i be type size_t.
for (size_t i = 0; i < n; i++) {
int e = name[i];
printf("The ASCII value of the letter %c is : %d \n", name[i], e);
}
Putting it all together, we get
#include <stdio.h>
#include <string.h>
int main() {
char name[100];
memset(name, 0, sizeof(name));
printf("Enter a name : \n");
fgets(name, sizeof(name), stdin);
size_t n = strlen(name);
if (n > 0 && name[n - 1] == '\n') {
name[n - 1] = '\0';
--n;
}
for (size_t i = 0; i < n; i++){
int e = name[i];
printf("The ASCII value of the letter %c is : %d \n", name[i], e);
}
/* To correctly print a size_t, use %zu */
printf("%zu\n", n);
/* In C99 main implicitly returns 0 if you don't add a return value
yourself, but it's a good habit to remember to return from functions. */
return 0;
}
Which should work pretty much as expected.
Additional notes:
This code should be valid C99, but I believe it's not valid C89. If you need to write to the older standard, there are several things you need to do differently. Fortunately, your compiler should warn you about those issues if you tell it which standard you want to use. C99 is probably the default these days, but older code still exists.
It's a bit inflexible to be reading strings into fixed-size buffers like this, so in a real situation you might want to have a way of dynamically increasing the size of the buffer as necessary. This will probably require you to use C's manual memory management functionality like malloc and realloc, which aren't particularly difficult but take greater care to avoid issues like memory leaks.
It's not guaranteed the strings you're reading are in any specific encoding, and C strings aren't really ideal for handling text that isn't encoded in a single-byte encoding. There is support for "wide character strings" but probably more often you'll be handling char strings containing UTF-8 where a single codepoint might be multiple bytes, and might not even represent an individual letter as such. In a more general-purpose program, you should keep this in mind.
If we need write a code to get ASCII values of all elements in a string, then we need to use "%d" instead of "%c". By doing this %d takes the corresponding ascii value of the following character.
If we need to only print the ascii value of each character in the string. Then this code will work:
#include <stdio.h>
char str[100];
int x;
int main(){
scanf("%s",str);
for(x=0;str[x]!='\0';x++){
printf("%d\n",str[x]);
}
}
To store all corresponding ASCII value of character in a new variable, we need to declare an integer variable and assign it to character. By this way the integer variable stores ascii value of character. The code is:
#include <stdio.h>
char str[100];
int x,ascii;
int main(){
scanf("%s",str);
for(x=0;str[x]!='\0';x++){
ascii=str[x];
printf("%d\n",ascii);
}
}
I hope this answer helped you.....😊
I am trying to understand string's behavior in C and it is bothering me since my following two code snippets result into different output:
(For the sake of this question, Let us assume user enters 12)
int main(void)
{
char L_Red[2];
char temp[] = "I";
printf("Enter pin connected to red: ");
scanf("%s", L_Red);
strcat(temp,L_Red);
printf("%s \n", temp);
return 0;
}
this yields: 12 as output (and not I12) Why ?
int main(void)
{
char L_Red[2];
printf("Enter pin connected to red: ");
scanf("%s", L_Red);
char temp[] = "I";
strcat(temp,L_Red);
printf("%s \n", temp);
return 0;
}
This yields: I12I (and not, I12) Why ?
I have read about string in C and as per my understanding, neither am I allocating temp any fixed size and changing it later to get these vague outputs nor am I using strings like the way they are not supposed to. Is there any other concept at play here ?
The array temp is an array of two characters (the 'I' and the string terminator '\0'). That's it. Attempting to append more characters to that array will write out of bounds and lead to undefined behavior.
You need to make sure that the destination array temp have enough space to fit its original content plus the string you want to append (plus the terminator).
Also, if you want to input more than one character for the "string" L_Red you need to increase its size as well.
I also recommend you use a limit in the format specifier so you can't write out of bounds:
char L_Red[3]; // Space for two characters, plus terminator
scanf("%2s", L_Red); // Read at most two characters of input
You are getting strange answers because your destination string (ie the first argument to strcat) is not long enough to handle both strings plus a null termination character. Also the length of L_Red is too short as it does not have enough space for the null termination character either.
Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 5 years ago.
Improve this question
So I'm trying to make it so that you can write text into a file until you make a newline or type -1. My problem is that when you write, it just keeps going until it crashes and gives the error "Stack around the variable "inputChoice" was corrupted".
I believe the problem is that the program doesn't stop accepting stdin when you want to stop typing (-1, newline) and that causes the error. I've tried with a simple scanf and it works, but you can only write a word. No spaces and it doesn't support multiple lines either. That's why I have to use fgets
Judging from your comments, I assume that there are some basic concepts in C
that you haven't fully understood, yet.
C-Strings
A C-String is a sequence of bytes. This sequence must end with the value 0.
Every value in the sequence represents a character based on the
ASCII encoding, for example the
character 'a' is 97, 'b' is 98, etc. The character '\0' has
the value 0 and it's the character that determines the end of the string.
That's why you hear a lot that C-Strings are '\0'-terminated.
In C you use an array of chars (char string[], char string[SOME VALUE]) to
save a string. For a string of length n, you need an array of dimension n+1, because
you also need one space for the terminating '\0' character.
When dealing with strings, you always have to think about the proper type,
whether your are using an array or a pointer. A pointer
to char doesn't necessarily mean that you are dealing with a C-String!
Why am I telling you this? Because of:
char inputChoice = 0;
printf("Do you wish to save the Input? (Y/N)\n");
scanf("%s", &inputChoice);
I haven't changed much, got very demotivated after trying for a while.
I changed the %s to an %c at scanf(" %c, &inputChoice) and that
seems to have stopped the program from crashing.
which shows that haven't understood the difference between %s and %c.
The %c conversion specifier character tells scanf that it must match a single character and it expects a pointer to char.
man scanf
c
Matches a sequence of characters whose length is specified by the maximum field
width (default 1); the next pointer must be a
pointer to char, and there must be enough room for all the characters
(no terminating null byte is added). The usual skip of
leading white space is suppressed. To skip white space first, use an explicit space in the format.
Forget the bit about the length, it's not important right now.
The important part is in bold. For the format scanf("%c", the function
expects a pointer to char and its not going to write the terminating '\0'
character, it won't be a C-String. If you want to read one letter and one
letter only:
char c;
scanf("%c", &c);
// also possible, but only the first char
// will have a defined value
char c[10];
scanf("%c", c);
The first one is easy to understand. The second one is more interesting: Here
you have an array of char of dimension 10 (i.e it holds 10 chars). scanf
will match a single letter and write it on c[0]. However the result won't be
a C-String, you cannot pass it to puts nor to other functions that expect
C-Strings (like strcpy).
The %s conversion specifier character tells scanf that it must match a sequence of non-white-space characters
man scanf
s
Matches a sequence of non-white-space characters; the next pointer must be a
pointer to the initial element of a character array that is long enough to
hold the input sequence and the terminating null byte ('\0'), which is added
automatically.
Here the result will be that a C-String is saved. You also have to have enough
space to save the string:
char string[10];
scanf("%s", string);
If the strings matches 9 or less characters, everything will be fine, because
for a string of length 9 requires 10 spaces (never forget the terminating
'\0'). If the string matches more than 9 characters, you won't have enough
space in the buffer and a buffer overflow (accessing beyond the size) occurs.
This is an undefined behaviour and anything can happen: your program might
crash, your program might not crash but overwrites another variable and thus
scrwes the flow of your program, it could even kill a kitten somewhere, do
you really want to kill kittens?
So, do you see why your code is wrong?
char inputChoice = 0;
scanf("%s", &inputChoice);
inputChoice is a char variable, it can only hold 1 value.
&inputChoice gives you the address of the inputChoice variable, but the
char after that is out of bound, if you read/write it, you will have an
overflow, thus you kill a kitten. Even if you enter only 1 character, it will
write at least 2 bytes and because you it only has space for one character, a kitten will die.
So, let's talk about your code.
From the perspective of an user: Why would I want to enter lines of text, possibly a lot of lines of text
and then answer "No, I don't want to save the lines". It doesn't make sense to
me.
In my opinion you should first ask the user whether he/she wants to save the
input first, and then ask for the input. If the user doesn't want to save
anything, then there is no point in asking the user to enter anything at
all. But that's just my opinion.
If you really want to stick to your plan, then you have to save every line and
when the user ends entering data, you ask and you save the file.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define BUFFERLEN 1024
void printFile () {
int i;
char openFile[BUFFERLEN];
FILE *file;
printf("What file do you wish to write in?\n");
scanf("%s", openFile);
getchar();
file = fopen(openFile, "w");
if (file == NULL) {
printf("Could not open file.\n");
return;
}
// we save here all lines to be saved
char **lines = NULL;
int num_of_lines = 0;
char buffer[BUFFERLEN];
printf("Enter an empty line of -1 to end input\n");
// for simplicity, we assume that no line will be
// larger than BUFFERLEN - 1 chars
while(fgets(buffer, sizeof buffer, stdin))
{
// we should check if the last character is \n,
// if not, buffer was not large enough for the line
// or the stream closed. For simplicity, I will ignore
// these cases
int len = strlen(buffer);
if(buffer[len - 1] == '\n')
buffer[len - 1] = '\0';
if(strcmp(buffer, "") == 0 || strcmp(buffer, "-1") == 0)
break; // either an empty line or user entered "-1"
char *line = strdup(buffer);
if(line == NULL)
break; // if no more memory
// process all lines that already have been entered
char **tmp = realloc(lines, (num_of_lines+1) * sizeof *tmp);
if(tmp == NULL)
{
free(line);
break; // same reason as for strdup failing
}
lines = tmp;
lines[num_of_lines++] = line; // save the line and increase num_of_lines
}
char inputChoice = 0;
printf("Do you wish to save the Input? (Y/N)\n");
scanf("%c", &inputChoice);
getchar();
if (inputChoice == 'Y' || inputChoice == 'y') {
for(i = 0; i < num_of_lines; ++i)
fprintf(file, "%s\n", lines[i]); // writing every line
printf("Your file has been saved\n");
printf("Please press any key to continue");
getchar();
}
// closing FILE buffer
fclose(file);
// free memory
if(num_of_lines)
{
for(i = 0; i < num_of_lines; ++i)
free(lines[i]);
free(lines);
}
}
int main(void)
{
printFile();
return 0;
}
Remarks on the code
I used the same code as yours as the base for mine, so that you can spot the
differences much quicker.
I use the macro BUFFERLEN for declaring the length of the buffers. That's
my style.
Look at the fgets line:
fgets(buffer, sizeof buffer, stdin)
I use here sizeof buffer instead of 1024 or BUFFERLEN. Again, that's my
style, but I think doing this is better, because even if you change the size
of the buffer by changing the macro, or by using another explicit size, sizeof buffer
will always return the correct size. Be aware that this only works when
buffer is an array.
The function strdup returns a pointer a pointer to a new string that
duplicates the argument. It's used to create a new copy of a string. When
using this function, don't forget that you have to free the memory using
free(). strdup is not part of the standard library, it conforms
to SVr4, 4.3BSD, POSIX.1-2001. If you use Windows (I don't use Windows,
I'm not familiar with the Windows ecosystem), this function might not be
present. In that case you can write your own:
char *strdup(const char *s)
{
char *str = malloc(strlen(s) + 1);
if(str == NULL)
return NULL;
strcpy(str, s);
return str;
}