This question already has answers here:
Segmentation fault on large array sizes
(7 answers)
Segmentation Fault - Large Array
(1 answer)
Closed 8 years ago.
I am currently creating a large array that looks like this:
unsigned char arr[35000][500];
I then try to write in 256 characters into the array like so:
for(i=0; i < 256; i++)
{
arr[i][0] = i;
}
When I do this, I get the following seg fault:
Program received signal SIGSEGV, Segmentation fault.
0x00000000004007e3 in main () at arr.c:41
41 arr[i][0] = i;
Any suggestions on why this is happening?
You probably have some stack overflow happening. Consider using dynamic memory allocation
Related
This question already has answers here:
Stack smashing detected
(10 answers)
How do I determine the size of my array in C?
(24 answers)
Closed 9 months ago.
What is problem with my code, that it shows "stack smashing detected"
Problem Statement:
Given an array, we have to find the smallest element in the array.
#include<stdio.h>
int main(){
int arr[20],i,j,c,x,num;
scanf("%d",&num);
for(x=0;x<num;x++){
scanf("%d",&arr[x]);
}
for(i=0;i<sizeof(arr)-1;i++){
if(arr[i]>arr[i+1]){
c=arr[i];
arr[i]=arr[i+1];
arr[i+1]=c;
}
}
printf("%d",*(arr+0));
return 0;
}
If the user-provided value num is greater than 20, your code will write to memory off the end of the array arr. This is undefined behaviour, and likely to cause a crash.
Two problems:
(1) int arr[20] can only hold twenty values, but you let the user put in any number.
(2) sizeof(arr) gives you the size in bytes, not the number of elements.
The compiler is able to detect one or both of these problems, and give you an error message telling you.
This question already has answers here:
What's the rationale for null terminated strings?
(20 answers)
Closed 3 years ago.
The 2d array:
char c[4][3]={{'a','b','c'},{'d','e','f'},{'g','h','i'},{'j','k','l'}};
As I wanted to get 'def' after running the program, I tried this code:
printf("%s\n",c[1]);
However, the result was 'defghijkl\262'. What's wrong with my code?
You can print def in two ways:
char c[4][4]={{'a','b','c','\0'},{'d','e','f','\0'},{'g','h','i','\0'},{'j','k','l','\0'}};
printf("%s\n",c[1]);
So, basically printf needs null termination to know where to stop printing
or you can print using a loop without null termination like:
char c[4][3]={{'a','b','c'},{'d','e','f'},{'g','h','i'},{'j','k','l'}};
for(int i = 0; i < 3; i++)
{
printf("%c", c[1][i]);
}
This question already has an answer here:
Segmentation Fault, large arrays
(1 answer)
Closed 4 years ago.
The following code, when compiled and run, gives me a segmentation fault. Why is this?
#include <stdio.h>
#include <limits.h>
int main(void)
{
int fat_array[INT_MAX];
return 0;
}
What you are requesting is to have about 2,147,483,647integer spaces allocated to you. Each integer is usually four bytes so that's 8,589,934,588 bytes which is 8 gigabytes of memory. This is likely above the allowed amount of memory a single process is allowed to reserve, and for good reason, so you get an error.
This question already has answers here:
Why int array[1000][1000] is memory issue in C program? [duplicate]
(1 answer)
Getting a stack overflow exception when declaring a large array
(8 answers)
Closed 5 years ago.
I'm using a pretty simple 2D array to store values (it's part of a mandelbrot set program).
int toBeWritten[xres][yres]; // xres and yres are calculated based on command line arguments
The 2D array works fine until my numbers get larger.
These, for example, work:
int toBeWritten[1024][1160];
int toBeWritten[2048][2321];
But when the size of the array grows to this:
int toBeWritten[4092][4637]; // the size I start getting seg faults
int toBeWritten[8192][9284]; // the largest size I want to get to
I get a seg fault if I try and access this array at any point after creating it.
Is it simply too big? Am I not allocating memory correctly?
If I can't make a 2D array this large, how could I store the values instead?
Thanks for any help!
This question already has answers here:
Why don't I get a segmentation fault when I write beyond the end of an array?
(4 answers)
Closed 7 years ago.
I've brushing up on my C coding and going over some old exercises I did a couple of years back. I came accros a situation I am almost 100% sure it should give a segmentation fault, but instead the program runs smoothly and terminates correctly. Why is that happening?
#include <stdio.h>
int main(void){
int vals[6] = {0,0,0,0,0,0};
vals[8]++; //This should not be ok!!?
printf("Done");
return 0;
}
The behaviour of vals[8] is undefined.
It's equivalent to *(vals + 8) which is dereferencing memory outside the bounds of the array.
A "segmentation fault" is one of many things that could happen. The compiler could also eat your cat.