i am new to c programming. i am getting Segmentation fault (core dumped) when i am
trying to print the string. please help.
#include <stdio.h>
#include <string.h>
int main()
{
char *ptr;
strcpy(ptr, "mystring");
printf( "%s\n", ptr);
return 0;
}
You haven't allocated any memory for your pointer to point at.
char array[MAX_LEN + 1];
char *ptr = array;
strncpy(ptr, "Cadence", MAX_LEN);
ptr[MAX_LEN] = '\0';
printf( "%s\n", ptr);
Please note that strncpy() can be a safer way to copy strings, since we specify the maximum number of characters to copy, which makes it harder to overrun the string and 'scribble' memory.
Update in response to comments: I've altered the above code to use a slightly safer pattern. You might also want to investigate strlcpy() (non-standard library).
When you declare char *ptr;, you allocate memory for a pointer to a char. But if you want to put a string inside the char, it will make an overflow.
Therefore, you have to allocate memory for your string :
char str[1024]; // which is the maximum string lenth that you will be able to put in str.
Furthemore, don't forget the null terminator (\0) that terminate every string and has the size of one char
Related
I just started to learn memory management in C, and I didn't understand something. I want to allocate memory to a buffer that holds 12 bytes. which is the exact size of Hello World! without null terminator.
Then I want to append a string to the current string with strcat, and of course I cannot do that because I will get core dumped error.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, char const *argv[])
{
char mystr[12] = "Hello World!";
# allocate memory to mystr?
char *ptr = (char*) malloc(13 * sizeof(char));
strcat(mystr, "Hello");
return 0;
}
So, I don't know how can I allocate memory to the mystr variable if malloc doesn't take any other arguments except the target size.
I don't know how can I allocate memory to the mystr variable if malloc doesn't take any other arguments except the target size.
It is not possible to allocate extra memory to an array. Instead, what you want to do is allocate a new block of memory, copying the original string into the beginning of that memory (strcpy), then append the rest (strcat):
char *p = (char*) malloc((12 + 5 + 1) * sizeof(char));
strcpy(p, myptr);
strcat(p, "Hello");
12 for the first string, plus 5 for the second, plus one for the null-terminator.
Of course, since you know the final size, you could also simply allocate a big enough array instead of using malloc (and you can also use memcpy, too).
The problem should be that a string in C always end with a NULL character (also noted '\0'), so your string is actually 13 characters long. (That character is always automatically added with string literals and serves at telling where the string stops, because a string doesn't have a fixed length.)
So the strcat tries to read the string Hello world! followed by garbage (since the null-terminator is not included in the string).
P.S.: the error is not the core dumped but the Segmentation fault that precedes it, and this tells you that you are trying to change something in a segment you are not supposed to change (or execute/read something you are not supposed to -- this is a security feature).
Edit: after modifying the string mystr, you also need to change the length you allocate (in the malloc: use 13 * sizeof(char), or more simply here in this case sizeof(mystr)).
P.S. 2: also comments in C are started by //, not # (those are preprocessor directives).
you cant change the size of the array. mystr has to be also dynamically allocated.
int main(int argc, char const *argv[])
{
const char *ptr = "Hello World!";
const char *ptr2 = "hello";
char *mystr = malloc(strlen(ptr)+1);
strcpy(mystr, ptr);
mystr = realloc(mystr, strlen(mystr) + strlen(ptr2) + 1);
strcat(mystr, ptr2);
return 0;
}
I am a beginner with c, and I am having a problem with scanf and strings.
here is an example I wrote of my problem.
#include <stdio.h>
#include <string.h>
int main(void)
{
char* string;
scanf("%s", &string);
if (strcmp(string, "Foo") == 0) //segmentation fault here
printf("Bar");
}
basically, this code compiles, but when I run it I get a segmentation fault in strcmp()
if I replace the "string" in that line with "&string" it works, but I get this error from the compiler
/usr/include/stdio.h:362:12: note: expected 'const char * __restrict__' but argument is of type 'char **'
which makes me think that this solution is not really ideal.
also If I declare string like this:
char string[100];
that works without any warnings, but that is also not ideal because I am not sure how large the string is going to be.
Is there a better solution I'm missing here, or are these my only options?
thank you.
char* string;
scanf("%s", &string);
string is not pointing to any valid memory location. Allocate memory using malloc to an array of characters and copy input to it. Make sure allocated memory has space for null termination character. Remember to free the memory to avoid leaks.
Just try that code
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main(void)
{
char* string;
string=(char *)malloc(3); /*allocate the memory to string cahr pointer(default pointer point to single byte and if you print pointer variable don't used & character)*/
scanf("%s", string);
if (strcmp(string, "Foo") == 0)
printf("Bar\n");
}
While declaring a char *. it will not having any memory location. so you have to allocate a
memory location before you use that variable.
char *p;
p=malloc(sieof(char) * size of string);
then you use scanf() function. it will work properly.
when we are accessing a unknown memory(ie unallocated memory). then it will through the segmentation fault
I have written the following program (it is given as an example in one of the best text books). When I compile it in my Ubuntu machine or at http://www.compileonline.com/compile_c_online.php, I get "segmentation fault"
The problem is with while( *p++ = *str2++)
I feel it is a perfectly legal program. Experts, please explain about this error.
PS: I searched the forum, but I found no convincing answer. Some people even answered wrong, stating that *(unary) has higher precedence than ++ (postfix).
#include <stdio.h>
#include <string.h>
int main()
{
char *str1= "Overflow";
char *str2= "Stack";
char *p = str1;
while(*p)
++p;
while( *p++ = *str2++)
;
printf("%s",str1);
return 0;
}
Thanks
str1 and str2 point to string literals. You aren't allowed to modify those. Even if you could, there isn't enough memory allocated for the string to hold the characters you're trying to append. Instead, initialize a sufficiently large char array from a string literal:
char str1[14] = "Overflow";
I feel it is a perfectly legal program.
Unfortunately, it is not. You have multiple, severe bugs.
First of all, you are creating pointers to string literals char *str1= "Overflow"; and then you try to modify that memory. String literals are allocated in read-only memory and attempting to write to them results in undefined behavior (anything can happen).
Then you have while(*p) ++p; which looks for the end of the string, to find out where to append the next one. Even if you rewrite the pointers to string literals into arrays, you don't have enough free memory at that location. You must allocate enough memory to hold both "Overflow" and "Stack", together with the string null termination.
You should change your program to something like this (not tested):
#include <stdio.h>
int main()
{
char str1[20] = "Overflow"; // allocate an array with enough memory to hold everything
char str2[] = "Stack"; // allocate just enough to hold the string "Stack".
char *p1 = str1;
char *p2 = str2;
while(*p1)
++p1;
while(*p1++ = *p2++)
;
printf("%s",str1); // should print "OverflowStack"
return 0;
}
Or of course, you could just #include <string.h> and then strcat(str1, str2).
Because you are crossing the string boundary of Overflow (str1) is why you are getting sigsegv.
str1 does not have enough memory allocated to accomodate beyond Overflow.
#include <stdio.h>
#include <string.h>
int main() {
char tab[2]={"12"};
FILE *outfile;
char *outname = "/home/dir/";
printf("%s", strcat(outname,tab));
outfile = fopen(strcat(outname,btab), "w");
if (!outfile) {
printf("There was a problem opening %s for writing\n", outname);
}
}
I have this error: Segmentation Fault.
How can I fix it?
At least two errors:
char tab[2] = {"12"};
You'd better use tab[3] or even better tab[] -- you need one extra char for the terminating NUL character.
Also,
char *outname = "etc...";
creates a constant string in the data segment of the executable -- it can't be overwritten, since strcat is using its first parameter to concatenate the two strings. So when strcat() tries to do so, it segfaults. Use
char outname[50]; // something big enough
strcpy(outname, "/home/dir");
instead.
outname is a string literal and string literals are not modifiable. Modifying a string literal is undefined behavior.
outname is Const pointer so once you have entered some thing in it, you can't modify it.
However if you want to copy things in it, make a char array of the size equal to tab[] array because here the size of string to be copied is known. Most of the time char pointers like OUTNAME are used when you are taking input from a user once and you don't know how long that input will be.
In your code,
char *outname = "/home/dir/";
outname is a string literal and hence when used with strcat, it does not have enough length to hold the concatenated string.This results in segmentation fault.
Same is the case had you declared it as below,
char outname[] = "/home/dir/";
The solution for this to declare the size of the outname big enough to hold the concatenated string.
char outname[80] = "/home/dir/";
Below is my code
#import <stdio.h>
#import <string.h>
int main(int argc, const char *argv[])
{
char *str = "First string";
char *str2 = "Second string";
strcpy(str, str2);
return 0;
}
It compiles just fine without any warning or errors, but when I run the code I get the error below
Bus error: 10
What did I miss ?
For one, you can't modify string literals. It's undefined behavior.
To fix that you can make str a local array:
char str[] = "First string";
Now, you will have a second problem, is that str isn't large enough to hold str2. So you will need to increase the length of it. Otherwise, you will overrun str - which is also undefined behavior.
To get around this second problem, you either need to make str at least as long as str2. Or allocate it dynamically:
char *str2 = "Second string";
char *str = malloc(strlen(str2) + 1); // Allocate memory
// Maybe check for NULL.
strcpy(str, str2);
// Always remember to free it.
free(str);
There are other more elegant ways to do this involving VLAs (in C99) and stack allocation, but I won't go into those as their use is somewhat questionable.
As #SangeethSaravanaraj pointed out in the comments, everyone missed the #import. It should be #include:
#include <stdio.h>
#include <string.h>
There is no space allocated for the strings. use array (or) pointers with malloc() and free()
Other than that
#import <stdio.h>
#import <string.h>
should be
#include <stdio.h>
#include <string.h>
NOTE:
anything that is malloc()ed must be free()'ed
you need to allocate n + 1 bytes for a string which is of length n (the last byte is for \0)
Please you the following code as a reference
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, char *argv[])
{
//char *str1 = "First string";
char *str1 = "First string is a big string";
char *str2 = NULL;
if ((str2 = (char *) malloc(sizeof(char) * strlen(str1) + 1)) == NULL) {
printf("unable to allocate memory \n");
return -1;
}
strcpy(str2, str1);
printf("str1 : %s \n", str1);
printf("str2 : %s \n", str2);
free(str2);
return 0;
}
str2 is pointing to a statically allocated constant character array. You can't write to it/over it. You need to dynamically allocate space via the *alloc family of functions.
string literals are non-modifiable in C
Your code attempts to overwrite a string literal. This is undefined behaviour.
There are several ways to fix this:
use malloc() then strcpy() then free();
turn str into an array and use strcpy();
use strdup().
this is because str is pointing to a string literal means a constant string ...but you are trying to modify it by copying .
Note : if it would have been an error due to memory allocation it would have been given segmentation fault at the run time .But this error is coming due to constant string modification or you can go through the below for more details abt bus error :
Bus errors are rare nowadays on x86 and occur when your processor cannot even attempt the memory access requested, typically:
using a processor instruction with an address that does not satisfy
its alignment requirements.
Segmentation faults occur when accessing memory which does not belong to your process, they are very common and are typically the result of:
using a pointer to something that was deallocated.
using an uninitialized hence bogus pointer.
using a null pointer.
overflowing a buffer.
To be more precise this is not manipulating the pointer itself that will cause issues, it's accessing the memory it points to (dereferencing).
Let me explain why you do you got this error "Bus error: 10"
char *str1 = "First string";
// for this statement the memory will be allocated into the CODE/TEXT segment which is READ-ONLY
char *str2 = "Second string";
// for this statement the memory will be allocated into the CODE/TEXT segment which is READ-ONLY
strcpy(str1, str2);
// This function will copy the content from str2 into str1, this is not possible because you are try to perform READ WRITE operation inside the READ-ONLY segment.Which was the root cause
If you want to perform string manipulation use automatic variables(STACK segment) or dynamic variables(HEAP segment)
Vasanth
Whenever you are using pointer variables ( the asterix ) such as
char *str = "First string";
you need to asign memory to it
str = malloc(strlen(*str))
None of the mentioned solution, worked for me as I couldn't find where the error was coming from. So, I simply deleted my node_modules and re-installed it. And the error disappeared; my code started working again