I have written this program in Linux. I assigned the character array size to 10. Upon entering up to 23 characters, the program showed no error. But on the 24th character, it showed a segmentation error. Please tell me why we can enter excess characters.
#include <stdio.h>
void main ()
{
char a[10];
scanf("%s", &a);
printf("%s", a);
}
First remove & from scanf.Like
scanf("%s", a);
char a[10] allocates the array on the stack for 10 char only. now when you enter characters as soon as you get out of bounds you are overwriting other useful things like the stack frame of the scanf call.
This out-of-bounds behavior is undefined by the C standard,
Related
I'm trying to write a C program for mad libs game:
#include <stdio.h>
#include <stdlib.h>
int main()
{
char color[20];
char Noun[20];
char celebrity[20];
printf("Enter the color");
scanf("%s", color);
printf("enter the Noun");
scanf("%s", Noun);
printf("enter the celebrity");
scanf("%s", celebrity);
printf('roses are %s\n', color);
printf('%s are blue\n', Noun);
printf('I love %s\n', celebrity);
return 0;
}
After I type in my input at the "shell> " prompt, gcc gives me the following error:
Segmentation fault (core dumped)
Why do I get this error?
You don't limit the size of the scanned data (so it can write out of bounds if the input is larger than the len 20 arrays)
None of your arrays are initialized, so
Since you don't verify the return value from scanf, if a scanf fails, you'll try to read from uninitialized arrays
The last three of your format strings for your printf calls are using ' as the delimiter, not ", so they're actually character literals (yes, it's weird that C allows this syntactically), and their numeric value is being interpreted as a pointer that likely points to pure garbage; replace the single quotes with double quotes (").
#4 would almost certainly cause a segfault if the code got that far, but #1 could conceivably cause a segfault before you got to that point (and #2 and #3 could combine to cause a segfault if #4 was fixed).
I am learning C, it's my first programming language. I don't understand this Error called "Buffer Overflow". My code is as follows:
#include <stdio.h>
int main()
{
char a[5];
gets(a);
printf ("%s",a);
return 0;
}
Now when I type more that 5 words it should end with five, shouldn't it? But its showing some buffer error and I have no idea what to do about it. please help me with this.
isn't that a[5] is the word limit of 5?
I'm very confused.
Sorry if it distrub you all and thanks in advance.
Actually, the limit is 4 characters, because a null terminator will be added to the end in order to form a valid string. This means that you need char a[6] if you want space for 5 characters.
Also, gets shouldn't be used for this exact reason. Instead, I would use scanf:
scanf("%5s", a);
This will tell it to read 5 characters at most, even if there are more.
With those changes, the program should look like this:
#include <stdio.h>
int main()
{
char a[6];
scanf("%5s", a);
printf ("%s",a);
return 0;
}
It's also possible to use fgets instead:
fgets(a, 5, stdin);
This question already has answers here:
Why does scanf ask twice for input when there's a newline at the end of the format string?
(7 answers)
Closed 5 years ago.
I am trying to simply get a user input string. The string is saved (scanf) when the person hits enter or the string hits the NULL character. Yet, when I run my program, and type in the string, it continues input until I type \n or \0 (whichever I have in my if statement. Here's my code:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define MSL 30
char intersperse(char);
char widen_stars(char);
int main(int argc, char *argv[])
{
char *p, *q, *str1[MSL], *str2[MSL];
if(str1[MSL+1] != '\0'){
printf("Please enter a string of maximum 30 characters: ");
scanf("%s\n", str1[MSL]);
}
printf("%s\n", str1[MSL]);
}
This is by no means finished code, i'm just in the process of writing my program when I snagged this annoying bug. The +1 on my MSL is to make sure the NULL character is read so my string doesn't yell at me, idk if it's necessary, but it's a precaution. Thank you answerers.
There are various problems with your code, not just a simple bug. I'll try to describe some of them here.
Turn on compiler warnings. For example, if your array is size 30, there is no element at index 30 (indexes would go from 0 to 29). Your compiler could have warned you about that and other problems.
Store a string in a char array or pointer, but not in a char pointers array. When you wrote:
char *str1[MSL]
You are creating an array of pointers to char, or an array of strings. However, since you're dealing with pointers to char as strings, you'd need to allocate space for them yourself. That's another concept. You probably meant to write an array of char like this:
char str1[MSL]
char *p, *q, str1[MSL], str2[MSL]; /* in your declaration */
Enter at maximum 29 characters. If your string holds 30 char, and you need 1 to mark the null at the end, you've got 29 remaining usable chars. Or… change MSL to 31.
Consider dropping the \n from your scanf format. It will make it read all whitespace (which is what \n represents in a format) waiting for the next non-whitespace. Thus, the input somestring<enter> for example won't be sent directly to your program until the next non-whitespace or EOF.
Give scanf an address. That could be just the array name (which translates to the first element address) or a pointer if you were working with them.
scanf("%s", str1);
scanf("%s", strptr); /* strptr is a char pointer pointing
to previously allocated space */
You can limit the length of what scanf reads. To be safe:
scanf("%29s", str1);
Don't use uninitialised data in comparisons. When you wrote if(str1[MSL+1] != '\0'), what did you expect str1 to contain if you had never stored anything yet?
int
main(int argc, char **argv)
{
char *p, *q, str1[31], str2[31]; /* a lot of unused variables */
printf("Please enter a string of maximum 30 characters: ");
scanf("%30s", str1);
printf("%s\n", str1);
return 0;
}
I'm a begginer in C; I made this mainly to test using fgets() and stop using scanf(), but as I was messing around with this code something weird happened. I compiled and ran this program:
int main() {
int i;
char c;
printf("Input an integer: ");
scanf("%i", &i);
printf("Number received: %i\n", i);
getchar(); // clear '\n' from input
printf("Input a character: ");
//scanf("%c", &c);
fgets(&c, 2, stdin);
printf("Integer: <%i>\n", i);
printf("Char: <%c>\n", c);
return 0;
}
Input:
12
r
Output:
Input an integer: 12
Number received: 12
Input a character: r
Integer: <0>
Char: <r>
So, I am curious as to why 'i' is being set to 0. Commenting fgets() and uncommenting scanf() makes the program work as intended. Thanks in advance.
The line
fgets(&c, 2, stdin);
claims that &c is a pointer to a char within an array of chars, with at least one valid char immediately after c. It is not, so your program has undefined behavior.
(So anything could happen, but it looks like what actually happened in your case is the read overflowed the c variable and also modified a byte of the i variable, which happened to be adjacent in memory.)
the minimum size for a buffer that will be used with fgets() is 3 characters. And it doesn't hurt to make it significantly longer, for instance 1024 characters.
As it is, the char 'c' will ALWAYS be overrun, resulting in undefined behavior. Remember, fgets() inputs right up to a newline (or 1 less than the length given as the second parameter. And the function always appends a NUL byte So, replace:
char c'
with
char c[1024];
then replace
fgets(&c, 2, stdin);
with
fgets( c, sizeof(c), stdin );
Notice no need for a '&' in the call to fgets() because the array 'c[]' will degrade to the address of the first byte of that array.
I am using Ubuntu 14.04 and gcc 4.8.4
#include<stdio.h>
void main()
{
char c[15];
printf("enter a character\n");
scanf("%15c",c);
printf("%s\n",c);
}
Output:
enter a character
qwertyuiopasdfghjkl
qwertyuiopasdfg
When I execute the program above with more than 15 characters as input, the output does not return garbage values. But when I execute the program below:
#include<stdio.h>
void main()
{
char c[5];
printf("enter a character\n");
scanf("%5c",c);
printf("%s\n",c);
}
I am giving input which is more than 5 characters. It returns output with garbage values.
The %c directive is for reading individual characters and character arrays, but not for those arrays specifically used as C strings. In particular, it does not store a null character after the last character transferred from the input.
Therefore, both of your scanf() calls are perfectly fine, but the subsequent printf() calls are not. In each case, the contents of the array designated by c are not null-terminated, therefore attempting to print the array via an %s directive will cause scanf() to read past the end of the array, producing undefined behavior. That the undefined behavior in one case seems reasonable to you and the undefined behavior in the other case does not is irrelevant.
I advise against working with unterminated string-like arrays as you are doing. It would be better to make the arrays large enough for a terminator, and to ensure that they are, in fact, terminated. But if you must print unterminated or possibly unterminated character arrays via printf(), then be sure to use the precision field of the directive to limit the number of characters that may be printed:
#include<stdio.h>
int main(void) {
char c[5];
printf("enter a character\n");
scanf("%5c", c);
printf("%.5s\n", c);
}
You are using the wrong format specifier in scanf for a string.
#include <stdio.h>
int main(void) {
char c[15];
printf("enter a string\n");
scanf("%14s", c); // change %c to %s
printf("%s\n", c);
return 0;
}
Note that the length limit is one less than the array size, to allow room for a nul terminator.
If your input is more than 5 characters then your program will displays 5 characters perfectly and then start displaying garbage value because while printing the string it will print upto '\0' character. And if '\0' is not present at last then it will print garbage.