This question already has answers here:
Is it possible to modify a string of char in C?
(9 answers)
Closed 3 years ago.
Im currently teaching myself c, and during exercising pointers use I bumped into this problem-
I'm trying to replace a substring in a string that is pointed to (meaning,not a char array), with a substring from another pointed string.
char *str1="I like pizza!";
char *str2="love";
printf("%s\n", str1);
for (int i=2, j=0; j<4; i++, j++) {
*(str1+i)=*(str2+j);
}
printf("%s\n", str1);
Result should be- the way I see it- output of "I like pizza", followed by output of "I love pizza".
Instead, I get a segfault (error 139).
I scraped the web for a solution but couldn't find what is the problem.
(I am aware that the for loop is not perfect, to say the least, but that's not the issue here).
Please help me out :)
Because those are pointers to a read-only part in you're program's binary. You can't change the content. Try this instead:
char str1[] = "I like pizza!";
char str2[] = "love"; // actually, this one can stay as a pointer as we're only reading
Now the strings are copied to the stack and the program works as intended.
The strings pointed to by str1 and str2 are read-only.
The behaviour on attempting to change their contents is undefined.
Use char str1[] = "I like pizza!"; &c. instead, which copies the read-only string to str1[], which you are then free to modify.
Related
This question already has answers here:
Assigning strings to arrays of characters
(10 answers)
Closed 7 years ago.
Well first i'm using language c i still a beginner.
char S[20];
S ="ewer" ;
is that correct.
Arrays (including strings) can't be assigned in C. For strings you need strcpy, or preferably strncpy:
#include <string.h>
...
char S[20];
strcpy(S, "ewer"); // strcpy is fine for this example
strncpy(S, "ewer", sizeof(S)); // strncpy is safer in general and
// should be preferred over strcpy
No, that won't work.
The variable S is an array, and you can't assign to arrays like that in C. The string "ewer" is represented as an array of characters terminated by the character '\0'. To copy it into the array, you need to use a function:
strcpy(S, "ewer");
That's a good question. In fact, if you'd wrote char* S instead, the example would worked. You might be confused by the fact, that arrays and pointers have many alike things — like [] operator.
But you must understand that array is different from a pointer. One of the main differences is that you couldn't increment an array, like ++myArr (this code okay for pointer, but not for an array). The other one you see in the question: you can not reassign an array variable to point another array. That is exactly what you're trying to do: you're assigning to the array variable S a pointer to a place with the text "ewer", that won't work.
Assuming that you wanted to assign to an array the text, you could do the following:
char S[] = "ewer";
Here you say to the compiler to allocate as much space on the stack, as the "ewer" text (plus the zero end character) holds, and copy that text there. Notice the empty braces [] — you don't even need to manually count symbols, it would be done for you by a compiler.
This question already has an answer here:
Is modification of string literals undefined behaviour according to the C89 standard?
(1 answer)
Closed 7 years ago.
So I've been playing around with C lately and have been trying to understand the intricacies of passing by value/reference and the ability to manipulate a passed-in variable inside a function. I've hit a road block, however, with the following:
void modifyCharArray(char *input)
{
//change input[0] to 'D'
input[0] = 'D';
}
int main()
{
char *test = "Bad";
modifyCharArray(test);
printf("Bad --> %s\n", test);
}
So the idea was to just modify a char array inside a function, and then print out said array after the modification completed. However, this fails, since all I'm doing is modifying the value of input that is passed in and not the actual memory address.
In short, is there any way I can take in a char *input into a function and modify its original memory address without using something like memcpy from string.h?
In short, is there any way I can take in a char *input into a function and modify its original memory address without using something like memcpy from string.h?
Yes, you can. Your function modifyCharArray is doing the right thing. What you are seeing is caused by that fact that
char *test = "Bad";
creates "Bad" in read only memory of the program and test points to that memory. Changing it is cause for undefined behavior.
If you want to create a modifiable string, use:
char test[] = "Bad";
This question already has answers here:
C segmentation fault-char pointers
(3 answers)
Closed 8 years ago.
I'm new in learning the C-Language and I have a question to pointers.
For Example if I try this:
char *pointer;
strcpy(pointer,"Hello, World!\n");
printf(pointer);
I get this Output:
Segmentation Fault
But if I try this:
char *pointer = "Hello, World!\n");
printf(pointer);
I get this:
Hello, World!
My question is why it isn't working with strcpy.
The functions do the same overall.
What is the difference between the first sourcecode and the second?
It would be good if someone could explain what is happening in the memory, so that I can get a better view at this.
char* pointer merely gives you a variable to use to access the memory location. You've not yet allocated any memory, so when you do the strcpy you're writing to whatever random/undefined value pointer has.
You need to do something like:
char* pointer = calloc(LEN);
if (pointer)
{
strcpy(pointer, "Hello World");
printf(pointer);
free(pointer);
}
The first parameter to "strcpy" should be pointing to a usable location in memory. In your first source code, you havent initialized "pointer" to anything. You should first initialize it, for example by declaring it to be an array of characters of maximum length:
char myArray[42]; // where 42 represents the maximum capacity
This question already has answers here:
Is the function strcpy always dangerous?
(9 answers)
Closed 8 years ago.
I am a beginner, and I am learning how to copy a string in C now.
Here is a problem I just met:
Every time I try to use "strcpy" command to copy from string 1 to string 2, Visual Studio 2013 will give me an error/warning message saying that "strcpy" is unsafe and suggest me to use strcpy_s instead.
Can you please explain why is strcpy unsafe to use? And what are the safer alternatives to strcpy?
Here is my code:
#include<stdio.h>
#include<string.h>
main()
{
char str1[] = "Copy a string.";
char str2[15];
char str3[15];
int i;
/* with strcpy() */
strcpy(str2, str1);
/* Without strcpy() */
for (i = 0; str1[i]; i++)
str3[i] = str1[i];
str3[i] = '\0';
/* Display str2 and str3 */
printf("The content of str2: %s\n", str2);
printf("The content of str3: %s\n", str3);
return 0;
}
P.S. I am using Visual Studio 2013 64-bit ultimate on Windows 7 64-bit ultimate.
Thank you for your time!
strcpy has no way of knowing how large the destination buffer is (i.e. there is no length parameter) so sloppy programming using it can lead to overrunning the buffer and corrupting other memory. Such an overrun can lead to crashes, odd behaviour and may be exploitable by malware authors.
BTW, look at strncpy that takes a length parameter. One problem to be aware of is that strncpy will not terminate the string if the buffer is too small so it is dangerous in its own way.
In answer to your comment/question - it will depend on context. If you know the buffer is large enough (for example, you allocated it in the previous line at the correct size), use strcpy. If there is any possibility of an overflow then use strncpy and make sure you set the last position in the buffer to null. Note that "any possibility" should be interpreted very generously - unless the previous five or six lines are enough to show safety then use strncpy.
Also see Andon's comment above about strncpy_s. Finally, if you do use strcpy, you might want to add a #pragma to suppress the warning at that location.
Here's your original code:
int main() {
char str1[] = "Copy a string.";
char str2[15];
strcpy(str2, str1);
}
Now, three day later, you go to edit this code. You've realized that you actually need to write the message "This code is copying a string." instead.
int main() {
char str1[] = "This code is copying a string.";
char str2[15];
strcpy(str2, str1);
}
However, you've now accidentally introduced a buffer overflow.
strcpy_s requires the extra argument of the length of str2. This way, while you may not get the whole string into your new buffer, you won't cause undefined behavior.
Unlike strncpy and strcpy_s, strcpy doesn't do any length checking of the destination buffer which may cause stack overflow allowing an attacker to execute malicious code if exploited or just crach your application
This question already has answers here:
Accessing an array out of bounds gives no error, why?
(18 answers)
Closed 9 years ago.
who can tell me why the code below still works? it is obvious that the str[4] is out of boundry:
#include<stdio.h>
int main(){
char str[3];
scanf("%s", str);
printf("%c\n", str[4]);
printf("%s\n", str);
return 0;
}
when it runs, enter abcdefg, it will echo the 5th char and the whole string, nothing
will be wrong, weird?
It has been declared that c/c++ doesn't do the boundary checking, while in the case above,
how should I use printf to print a c-string that the user has entered? or more generally, how to properly use a c-string that comes from users?
str[4] gives you a pointer to the memory address after the last element of your string. You can still convert this to a character, but you never know what you get and your software might crash.
Why does it work?
It doesn't "work". It might appear to be working, though; since your code (accessing an array out of bounds) invokes undefined behavior, you can get literally any result. Undefined behavior doesn't mean "it must crash".
You wrote:
when it runs, enter [abcdefg], it will echo the 5th char and the whole
string, nothing will be wrong, weird?
but I see no input reading in the code you posted:
#include<stdio.h>
int main(){
char str[3];
printf("%c\n", str[4]);
printf("%s\n", str);
return 0;
}
In any case, str[4] points to the 5-th char (byte) starting from str, so it will print any "garbage" that happens to be in that memory location when your program runs.
Accessing array items out of bounds is undefined behavior, so the exact behavior is not specified.
Just pay attention to your array bounds.