This question already has answers here:
What is the difference between char s[] and char *s?
(14 answers)
Closed 10 years ago.
I need to swap two characters by pointers but when I run this code,the program crashes.
int main(){
char *s1 = "string1";
swap(st,(st+1));
/* BUT THIS CODE WORKS - Whats the problem?
* char s1[] = "string1";
* swap(s1,&s1[1]);
*/
return 0;
}
void swap(char * const ptr1, char * const ptr2){
char temp = *ptr1;
*ptr1 = *ptr2;
*ptr2 = temp;
}
char *s1 = "string1";
Because s1 points to a string literal and modifying invokes undefined behaviour in C. That's why this doesn't work.
Whereas in this char s1[] = "string1";
s1 is an array and hence it can be modified.
A string literal may not be modified. You try to modify "string1" in your code which is not allowed. Indeed, many compilers place string literals in a special section that may not be written into.
This line of code creates a string literal which cannot be changed. It is only readable.
char *s1 = "string1";
Any attempt to change it will give you an error.
While your commented example :
char s1[] = "string1";
creates an actual array. This can be edited and used normally.
Related
This question already has answers here:
Assigning strings to arrays of characters
(10 answers)
Closed 3 years ago.
I want to store a string "hello" in an array of characters char arr[10]
when I run the following code:
#include <stdio.h>
int main(void)
{
char arr[10] = "Hello";
printf("%s", arr);
}
The output is generated fine.
But when this code below is run, I get an error
#include <stdio.h>
int main (void)
{
char arr[10];
char * str = "Hello";
arr = str;
printf("%s", arr);
}
Why is that in the first case the string can be stored in the array and not in the second case?
Is there a way I can store "hello" in the second case?
You should use strncpy() or even better strlcpy() if you are on a BSD system or macOS and portability is not the main concern.
In C, char *p means a pointer to a character string. While writing the following
char arr[10];
char * str = "Hello";
arr = str;
you might have thought that the character string stored in the memory location pointed by str would be copied over to the buffer arr, but C does not do this for you.
The code below does what you want
#include <stdio.h>
#include <string.h>
#define BUFSIZ 10
int main(){
char arr[BUFSIZ];
const char *str=“Hello”; /* I used const here */
strncpy(arr, str, BUFSIZ);
printf(“%s\n”, arr);
return 0;
}
Use strncpy or strlcpy instead of strcpy. Documentation for strncpy is here.
The strcpy is the way to do it, as noted above. You get an error for
arr = str;
in your given code because you are attempting to assign an array, and C does not allow arrays to be assigned. In your first part, the declaration you give
char arr[10] = "Hello";
is legal because it is considered an initialization (since it's in a declaration), not an assignment. That is allowed.
C is not always a bundle of consistency.
To copy a string in C, use strcpy:
char arr[10];
char * str = "Hello";
strcpy(arr, str);
printf("%s\n", arr);
Don't forget to #include <string.h>, which is required for you to have access to this function.
char arr[10] means that arr can be changed;
char * str = "Hello" equals to const char* str = "Hello", so it can not be changed.
so arr = str will occur error.
You should check the memory allocation mechanism in C.
This is initialization of char array; there are special rules
char x[] = "foobar";
This is assignment to a pointer
char *p;
p = "foobar"; // char *p = "foobar"; initialization of pointer
There is no direct way to do assignment of a value to an array of char. You need to assign each individual element
char x[7];
x[0] = 'f'; x[1] = 'o'; x[2] = 'o';
x[3] = 'b'; x[4] = 'a'; x[5] = 'r'; x[6] = 0;
or use a function to do that for you
strcpy(x, "foobar");
I am trying to perform swap operation as shown in below program, but it seems to crash when I am copying the element b into a (line 3 in swap() method) - this happens if the input is a char string i.e, char *str; If it is a character array instead (as in char str[];) the swap operation works fine.
#include<stdio.h>
void swap(char *a, char *b)
{
char tmp;
tmp = *a;
*a = *b; /* crash occurs here */
*b = tmp;
}
void func(char *a)
{
swap(a+1, a+2);
}
int main()
{
char *str = "abc";
// char str[] = "abc"; /* with this char array, the above swap func works */
func(str);
}
I think this is related to some C string rules that I seem to be unaware of. Please help!
String literals are read-only, trying to modify a string literal leads to undefined behavior.
Simplest solution? Declare it as an array instead:
char str[] = "abc";
char *str = "abc";
Above puts the string in the constant data section (also known as .rdata) of the program.As this is treated as constant data, it can not be modified
char str[] = "abc";
Above puts the string in the stack area of the program, as declared inside the function scope.This data can be modified.
As this is about the storage of string data, you are getting Crash in first declaration.
'char *str = "abc";'
The above places the abc in read only memory. So it cannot be modified.
'char str[]="abc";' is placed in stack. so it can be modified.
This question already has answers here:
Why does writing to a string literal in this C program segfault?
(6 answers)
Why can't I write to a string literal while I *can* write to a string object?
(4 answers)
Closed 8 years ago.
Why do I get exception in the code?
char* first = "first";
char* second = "second";
*first = *second;
Shouldn't it just assign values?
Error message says: access violation
No because they are string literals and are read only
char *first = "first";
char *second = "second";
you can try with arrays instead.
char first[] = "first";
char second[] = "second";
to prevent this kind of error you can do this
const char *first = "first";
const char *second = "second";
when declaring string literals, it will not prevent the problem completely because you still can cast away the const but you should do so conciously.
This question already has answers here:
Why can't I use this code to overwrite a string?
(4 answers)
Closed 9 years ago.
I am totally new to C and I am about to write a function that reverses a string. My code looks like this:
char *str = "abcdef";
char *ptr;
for(ptr = str; *ptr ; ptr++);
for(; str < --ptr; str++)
{
char c = *str;
*str = *ptr;
*ptr = c;
}
I get a segmentation fault error. And I don't see the mistake (maybe it is too obvious). Any hints?
Cheers
Change
char *str = "abcdef";
to
char str[] = "abcdef";
The first str points to a string literal and string literals are not modifiable in C,
Your string is stored in ROM, so you can't write to it. Depends on your platform though.
This question already has answers here:
How to replace specific characters in a string with other characters
(3 answers)
Closed 9 years ago.
I am trying to do something really basic on C but I keep getting a segmentation fault. All I want to do is replace a letter of a word with a different letter- in this example replace the l with a L. Can anyone help explain where I have gone wrong? It should be a really basic problem I think, I just have no idea why its not working.
#include<stdio.h>
#include<stdlib.h>
int main(int argc, char *argv[])
{
char *string1;
string1 = "hello";
printf("string1 %s\n", string1);
printf("string1[2] %c\n", string1[2]);
string1[2] = 'L';
printf("string1 %s\n", string1);
return 0;
}
For my output I get
string1 hello
string1[2] l
Segmentation fault
Thanks!
string1 = "hello";
string1[2] = 'L';
You can't change string literals, it's undefined behavior. Try this:
char string1[] = "hello";
Or maybe:
char *string1;
string1 = malloc(6); /* hello + 0-terminator */
strcpy(string1, "hello");
/* Stuff. */
free(string1);
char *string1;
string1 = "hello";
string1 points to a string literal and string literals are non-modifiable.
What you can do is initialize an array with the elements of a string literal.
char string1[] = "hello";
the elements of string1 array are modifiable.
char *string1 = "hello";
When running the code, the string literal will be in a section that is read-only. OS does not allow the code to change that block of memory, so you get a seg-fault.
char string1[] = "hello";
The string literal will be pushed onto the stack when you run the code.
string1[2] = 'L';
you are trying to change a string literal which is undefined behavior in C.
Instead use string1[]="hello";
Segmentation fault you get is because the literal is probably stored in the the read only section of the memory and trying to write to it produces undefined behavior.