This question already has answers here:
Crash or "segmentation fault" when data is copied/scanned/read to an uninitialized pointer
(5 answers)
Closed 6 years ago.
this is my code giving segmentation fault
#include <stdio.h>
#include <signal.h>
#include <stdlib.h>
#include <unistd.h>
#include <string.h>
int main(void) {
char *get;
scanf("%s", get);
int k = strcmp("sachin", get);
printf("%d", k);
}
thanks for any help;
char *get;
The above statement defines get to be a pointer to a character. It can store the address of an object of type char, not a character itself. The problem is with both scanf and strcmp call. You need to define an array of characters to store the input string.
#include <stdio.h>
#include <string.h>
int main(void) {
// assuming max string length 40
// +1 for the terminating null byte added by scanf
char get[40+1];
// "%40s" means write at most 40 characters
// into the buffer get and then add the null byte
// at the end. This is to guard against buffer overrun by
// scanf in case the input string is too large for get to store
scanf("%40s", get);
int k = strcmp("sachin", get);
printf("%d", k);
return 0;
}
You need to allocate memory for the pointer get.
Or use a char array:
char get[MAX_SIZE];
scanf("%s",get);
When you declare a pointer, it is not pointing to any memory address. You have to explicitly allocate memory for that pointer by malloc (c-style) or new (c++ style).
You will also need to manage the clean operation yourself by free / delete respectively.
of sure you get a segment fault.
you told your computer to write user input to uninitialized pointer.
char *get; // just a pointer that may wildly points anywhere
scanf("%s",get);
define get to be some char array
char get[SOME_LARGE_CONSTANTS_THAT_CLEARLY_IS_LARGER_THAN_YOUR_USER_INPUT];
Related
This question already has answers here:
Why do I get a segmentation fault when writing to a "char *s" initialized with a string literal, but not "char s[]"?
(19 answers)
Closed last month.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
main()
{
char *str;
int len;
printf("Enter the expression: \n");
scanf("%[^\n]", &str);
printf("%s\n", str);
len = strlen(str);
printf("%d\n", len);
}
I am trying input a string into a string pointer but it keeps giving me a segmentation error, however when i initialize it as char array it works fine.
It looks like you are trying to get scanf to allocate the necessary memory for the string. That option is only available as an extension in some implementations, but here's how that would work:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void) { // note the proper declaration
char *str = NULL;
int len;
printf("Enter the expression: \n");
if(scanf("%m[^\n]", &str) == 1) { // add `m` to malloc memory for the string
printf("%s\n", str);
len = strlen(str);
printf("%d\n", len);
free(str); // and `free` it after use
}
}
This is because you are using the scanf function to write to an uninitialized pointer str. The & is also uneeded for passing a pointer for scanf. A better way to handle the input would be to use fgets() or getline() and allocating memory dynamically using malloc() to store the input string.
When I use the whole code in the main function it works perfectly but now I want to use the function for some "Strings" which I initialize in a 2D-Array.
The idea behind the function is to create a product of a struct globally initialized. The line with strcpy gives the error:
Thread 1: EXC_BAD_INSTRUCTION (code=EXC_I386_INVOP, subcode=0x0).
I am using Xcode 11.3.1 on a Mac.
Can you guys help me with that?
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <time.h>
struct produkt neues_angebot(char *produktname);
struct produkt{
char name[5];
int produkt_zahl;
float produkt_preis;
};
struct produkt neues_angebot(char *produktname){
time_t t;
srand((unsigned) time(&t));
struct produkt Zwischenname = {
"xxx",(rand() % 49),((rand()% 600)/100)
};
strcpy(Zwischenname.name, produktname);
return Zwischenname;
}
int main(int argc, const char * argv[]) {
char teste[]="hello";
printf("%s\n",neues_angebot(teste).name);
}
Strings in C are null terminated. The string "hello" uses 6 bytes of storage: the 5 characters h e l l o plus the terminating null character.
Now strcpy does no length checking, so strcpy(Zwischenname.name, produktname) will attempt to copy all 6 of those bytes into the array Zwischenname.name, which only has space for 5 bytes. That overflows the array and is undefined behavior - the crash you observed is one of the many possible results.
The strcpy will copy the last character of teste, the null terminator '\0', to outside the bounds of the the destination buffer as it is one byte shorter than it needs to be, this will cause Zwischenname.name to not be null terminated.
The problem is printf will need the null-terminator to know where the string ends in order work correctly, passing a non null-terminated character array as an argument to printf("%s", argument); invokes undefined behavior.
Remember, in order for an array of characters to be treated as a propper string it needs to be null terminated, many library fuctions in C rely on this principle. Always make sure that it is.
So I'm trying to reverse a string, but I get a memory fault. Memory for s and s1 is initialized enough to accomodate the '/0' character as well.
Interestingly if I remove *s=*s1 and print s1 instead the program works.
But I haven't even set the "\0" character at the end of s1 so how does it even know where to stop printing?
And in the case below what exactly is the issue?
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
void main(void)
{
char *s = "abcdefghijklmnop", *s1=malloc(17);
int i;
for(i=0;i<strlen(s);i++)
{
*(s1+i) = *(s+strlen(s)-1-i);
}
*s=*s1;
printf("%s",s);
}
The char *s = "abcdefghijklmnop" is a string literal which is often in read-only memory.
An error will be generated if you attempt to modify a string literal.
You attempt to replace the first character in s with the first character in s1 when you do *s=*s+1.
If s wasn't a string literal, you should've done s=s1 instead of *s=*s1 to make s its reverse.
More about this:
https://softwareengineering.stackexchange.com/questions/294748/why-are-c-string-literals-read-only
String literals: Where do they go?
where in memory are string literals ? stack / heap?
The correct string is printed with printf("%s", s1); even if no \0 was stored because the memory next to the last character just happened to be 0 which is equivalent to \0. This needn't always be so and cannot be relied upon as malloc() doesn't initialise the memory that it allocates.
But calloc() will initialise the memory it allocates to 0.
See more here.
In your code *s=*s1 only copy the first content of s1 into s. i.e, *(s+0)=*(s1+0) not the entire string. So we need to assign the address of s1 to s. i.e, s=s1.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main()
{
char *s = "abcdefghijklmnop", *s1=malloc(17);
int i;
for(i=0;i<strlen(s);i++)
{
*(s1+i) = *(s+strlen(s)-1-i);
}
s=s1;
printf("%s",s);
free(s1);
return 0;
}
It is good practice to free the memory after its use.
So i've make a program about an char array that is dinamically allocated, the program is not yet finished but i've got some different return values after i run the program, no warnings, no errors in compiler so i don't know why i get this also sometimes the program crashes..
#include <stdlib.h>
#include <stdio.h>
#include <malloc.h>
#include <string.h>
int main(){
int n,i;
char *tab=NULL;
printf("New size of array: ");
scanf("%d",&n);
tab=(char*)malloc(n*sizeof(char));
memset(tab,'X',n*sizeof(tab));
for(i=0;i<n;i++)
printf("%c",tab[i]);
free(tab);
return 0;
}
In your memset you write n * sizeof(tab), I think you wanted to write :
n * sizeof(char)
You can also add a +1 add the end of your malloc and check the return value, just for security.
The C library function void *memset(void *str, int c, size_t n) copies the character c (an unsigned char) to the first n characters of the string pointed to, by the argument str.
memset(tab,'X',n*sizeof(char));
You've written n*sizeof(tab), you are copying the character X to unknown memory which might cause crash during runtime.
Look into this line:
memset(tab,'X',n*sizeof(tab));
You declared tab as a pointer, and a storage size of a pointer is more than one byte. Multiplying that by n which is the size of the array you want will cause you to copy data to memory space you're not allowed to access which is why you get a segmentation fault.
If you must use sizeof, then do this:
memset(tab,'X',n*sizeof(char));
or better yet, just do this:
memset(tab,'X',n);
because a char is one byte and n times 1 = n.
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