What is the problem with the following code? C compiler shows me error: Segmentation fault.
#include <stdio.h>
#include <math.h>
int main() {
int n = 4;
float A[n][n][n][n];
A[n][n][1][1] = 1;
A[n][n][1][2] = 0;
A[n][n][1][3] = -3;
A[n][n][1][4] = 0;
A[n][n][2][1] = 1;
A[n][n][2][2] = 0;
A[n][n][2][3] = -3;
A[n][n][2][4] = 0;
return 0;
}
All these assignments access memory outside the defined variable length array.
A[n][n][1][1]=1;
A[n][n][1][2]=0;
A[n][n][1][3]=-3;
A[n][n][1][4]=0;
A[n][n][2][1]=1;
A[n][n][2][2]=0;
A[n][n][2][3]=-3;
A[n][n][2][4]=0;
That is the valid range of indices for each dimension of the array is [0, n ). Thus the expression A[n] is invalid.
It seems you mean the following assignments
A[0][0][0][0]=1;
A[0][0][0][1]=0;
A[0][0][0][2]=-3;
A[0][0][0][3]=0;
A[0][0][1][0]=1;
A[0][0][1][1]=0;
A[0][0][1][2]=-3;
A[0][0][1][3]=0;
If you define an array of length N, the access to the last element is at the N-1 index, and the access to the first element is at index 0 because indexes start from 0 and end at N-1.
So your code should be like this:
#include<stdio.h>
#include<math.h>
#define N 4
int main() {
float A[N][N][N][N];
A[N-1][N-1][0][0]=1;
A[N-1][N-1][0][1]=0;
A[N-1][N-1][0][2]=-3;
A[N-1][N-1][0][3]=0;
A[N-1][N-1][1][0]=1;
A[N-1][N-1][1][1]=0;
A[N-1][N-1][1][2]=-3;
A[N-1][N-1][1][3]=0;
return 0;
}
Also, you're using a variable to define the length of your array, which is good only in the C99 standard and onwards. If you don't need a VLA (variable length array), you should use the #define instead.
int n=4;
float A[n][n][n][n];
A[n][n][1][1]=1;
is not valid in every C standard (read n1570 or better). You may want to code const int n=4; or #define n 4
(on cheap microcontrollers you might not even have enough stack space for 256 floats -> possible stack overflow)
And then, you always have a buffer overflow (read about undefined behavior). In A[x][x][1][1] the valid indexes are from x being 0 to x being 3 -that is n-1 - (not 4).
You might want to code
A[n-1][n-1][1][1] = 1;
but you should know that A[0][0][0][0] is being uninitialized. It may contain garbage (signalling NaN), and an hypothetical printf("%f\n", A[0][0][0][0]); could fail ....
I guess that Frama-C would have caught your mistake (I leave you to check that). Be aware of Rice's theorem.
I would initialize A with zeros (to ease debugging, and to have more reproducible runs) -after having corrected the buffer overflow-:
float A[n][n][n][n]={0.0};
Related
I'm writing this code in C for some offline games but when I run this code, it says "runtime failure #2" and "stack around the variable has corrupted". I searched the internet and saw some answers but I think there's nothing wrong with this.
#include <stdio.h>
int main(void) {
int a[16];
int player = 32;
for (int i = 0; i < sizeof(a); i++) {
if (player+1 == i) {
a[i] = 254;
}
else {
a[i] = 32;
}
}
printf("%d", a[15]);
return 0;
}
Your loop runs from 0 to sizeof(a), and sizeof(a) is the size in bytes of your array.
Each int is (typically) 4-bytes, and the total size of the array is 64-bytes. So variable i goes from 0 to 63.
But the valid indices of the array are only 0-15, because the array was declared [16].
The standard way to iterate over an array like this is:
#define count_of_array(x) (sizeof(x) / sizeof(*x))
for (int i = 0; i < count_of_array(a); i++) { ... }
The count_of_array macro calculates the number of elements in the array by taking the total size of the array, and dividing by the size of one element.
In your example, it would be (64 / 4) == 16.
sizeof(a) is not the size of a, but rather how many bytes a consumes.
a has 16 ints. The size of int depends on the implementation. A lot of C implementations make int has 4 bytes, but some implementations make int has 2 bytes. So sizeof(a) == 64 or sizeof(a) == 32. Either way, that's not what you want.
You define int a[16];, so the size of a is 16.
So, change your for loop into:
for (int i = 0; i < 16; i++)
You're indexing too far off the size of the array, trying to touch parts of memory that doesn't belong to your program. sizeof(a) returns 64 (depending on C implementation, actually), which is the total amount of bytes your int array is taking up.
There are good reasons for trying not to statically declare the number of iterations in a loop when iterating over an array.
For example, you might realloc memory (if you've declared the array using malloc) in order to grow or shrink the array, thus making it harder to keep track of the size of the array at any given point. Or maybe the size of the array depends on user input. Or something else altogether.
There's no good reason to avoid saying for (int i = 0; i < 16; i++) in this particular case, though. What I would do is declare const int foo = 16; and then use foo instead of any number, both in the array declaration and the for loop, so that if you ever need to change it, you only need to change it in one place. Else, if you really want to use sizeof() (maybe because one of the reasons above) you should divide the return value of sizeof(array) by the return value of sizeof(type of array). For example:
#include <stdio.h>
const int ARRAY_SIZE = 30;
int main(void)
{
int a[ARRAY_SIZE];
for(int i = 0; i < sizeof(a) / sizeof(int); i++)
a[i] = 100;
// I'd use for(int i = 0; i < ARRAY_SIZE; i++) though
}
I was trying to make an array that contains Fibonacci numbers in C, but I got into trouble. I can't get all of the elements, and some of the elements are wrongly calculated, and I don't know where I am I going wrong.
#include <stdio.h>
int main(void){
int serie[]={1,1},sum=0,size=2;
while(size<=4000000){
serie[size]=serie[size-1]+serie[size-2];
printf("%d\n",serie[size-1]);
size+=1;
}
return 0;
}
Output:
1
2
4
6
11
17
28
45
73
118
191
309
500
809
1309
2118
3427
5545
8972
14517
23489
38006
61495
99501
160996
260497
421493
681990
1103483
1785473
2888956
4674429
7563385
12237814
19801199
32039013
51840212
83879225
135719437
219598662
355318099
574916761
930234860
1505151621
-1859580815
-354429194
2080957287
1726528093
-487481916
1239046177
751564261
1990610438
-1552792597
437817841
-1114974756
-677156915
-1792131671
1825678710
33547039
1859225749
1892772788
-542968759
1349804029
806835270
-2138327997
-1331492727
825146572
-506346155
318800417
-187545738
131254679
-56291059
74963620
18672561
93636181
112308742
205944923
318253665
524198588
842452253
1366650841
-2085864202
-719213361
1489889733
770676372
-2034401191
-1263724819
996841286
-266883533
729957753
463074220
1193031973
1656106193
-1445829130
210277063
-1235552067
-1025275004
2034140225
1008865221
-1251961850
-243096629
-1495058479
-1738155108
1061753709
-676401399
385352310
-291049089
94303221
-196745868
-102442647
-299188515
-401631162
-700819677
-1102450839
-1803270516
1389245941
-414024575
975221366
561196791
1536418157
2097614948
-660934191
--------------------------------
Process exited after 2.345 seconds with return value 3221225477
Press any key to continue . . .
I don't understand why it is giving that output.
int serie[]={1,1}
Declares an array of two elements. As the array has two elements and indices start from zero, it has valid indices - 0 and 1, ie. serie[0] is the first element and serie[1] is the second element.
int size=2;
while(..) {
serie[size]= ...
size+=1;
}
As size starts 2, the expression serie[2] = is invalid. There is no third element in the array and it writes to an unknown memory region. Executing such an action is undefined behavior. There could be some another variable there, some system variable, or memory of another program or it can spawn nasal demons. It is undefined.
If you want to store the output in an array, you need to make sure the array has enough elements to hold the input.
And a tip:
int serie[4000000];
may not work, as it will try to allocate 40000000 * sizeof(int), which assuming sizeof(int) = 4 is 15.2 megabytes of memory. Some systems don't allow to allocate that much memory on stack, so you should move to dynamic allocation.
You're having an integer overflow because the int size is ,at a certain leverl, not big enough to hold the numbers, so the number is wrapping round the size and giving false values.
Your program should be like:
#include <stdio.h>
int main(void){
long long unsigned series[100] = {1,1};
int size = 2;
while(size < 100){
series[size] = series[size-1] + series[size-2];
printf("%llu\n", series[size-1]);
size += 1;
}
return 0;
}
Although, size of long long unsigned is also limited, at a certain level, with such very big numbers in Fibonacci. So this will result in more correct numbers printed, but also will overflow at a certain level. It will overflow when the number exceeds this constant ULLONG_MAX declared in limits.h.
The problem with this code:
#include <stdio.h>
int main(void){
int serie[]={1,1},sum=0,size=2;
while(size<=4000000){
serie[size]=serie[size-1]+serie[size-2];
printf("%d\n",serie[size-1]);
size+=1;
}
return 0;
}
... is that it attempts to store a very long series of numbers (4 million) into a very short array (2 elements). Arrays are fixed in size. Changing the variable size has no effect on the size of the array serie.
The expression serie[size]=... stores numbers outside the bounds of the array every time it's executed because the only legal array index values are 0 and 1. This results in undefined behavior and to be honest you were lucky only to see weird output.
There are a couple of possible solutions. The one that changes your code the least is to simply extend the array. Note that I've made it a static rather than automatic variable, because your implementation probably won't support something of that size in its stack.
#include <stdio.h>
int serie[4000000]={1,1};
int main(void){
int size=2;
while(size<4000000){ // note strict less-than: 4000000 is not a valid index
serie[size]=serie[size-1]+serie[size-2];
printf("%d\n",serie[size-1]);
size+=1;
}
return 0;
}
The more general solution is to store the current term and the two previous terms in the series as three separate integers. It's a little more computationally expensive but doesn't have the huge memory requirement.
#include <limits.h>
#include <stdio.h>
int main(void)
{
int term0=0, term1=1, term2;
while(1)
{
if (term0 > INT_MAX - term1) break;// overflow, stop
term2 = term0 + term1;
printf("%d\n",term2);
term0 = term1;
term1 = term2;
}
return 0;
}
This also has the benefit that it won't print any numbers that have "wrapped around" as a result of exceeding the limits of what can be represented in an 'int`. Of course, you can easily choose another data type in order to get a longer sequence of valid output.
You have two problems:
You need to allocate more space in serie, as much as you are going
to use
Eventually the fib numbers will become too big to fit inside an integer, even a 64bit unsigned integer (long long unsigned), i think 90 or so is about max
See the modified code:
#include <stdio.h>
// Set maximum number of fib numbers
#define MAX_SIZE 90
int main(void) {
// Use 64 bit unsigned integer (can't be negative)
long long unsigned int serie[MAX_SIZE];
serie[0] = 1;
serie[1] = 1;
int sum = 0;
int size = 0;
printf("Fib(0): %llu\n", serie[0]);
printf("Fib(1): %llu\n", serie[1]);
for (size = 2; size < MAX_SIZE; size++) {
serie[size] = serie[size-1] + serie[size-2];
printf("Fib(%i): %llu\n", size, serie[size]);
}
return 0;
}
As you are only printing out the numbers, you don't actually have to store all of them
(only the two previous numbers), but it really doesn't matter if there's only 90.
I was tried to print values of a float array with 7 elements. I assigned only 3 values and they are belong to float, double and integer.
Code:
#include <stdio.h>
int main(){
float array [7];
float f = 3.24;
double d = 23.5;
int i = 4;
array[0] = f;
array[1] = i;
array[2] = d;
int n = sizeof(array)/sizeof(float);
printf("Number of Elements : %d \n\n\n",n);
for(int j = 0; j < n ; j++){
printf("%.2f ,",array[j]);
}
printf("\b ");
}
I got an output as follows :
Number of Elements : 7
3.24 ,4.00 ,23.50 ,-1.#R ,96627196995476105000000000000000.00 ,96629547147269436000000000000000.00 ,0.00
I want to clear is my code correct? And why last four values are different to each other. What are these values?
The last values are known as garbage values in C.
If you do not initialize a variable explicitly in C, it's value could be anything before you explicitly assign something to it. This anything could be garbage; the language standard does not specify what it should be.
You can read this blog for How C compiler decides garbage values: http://blog.aditech.info/2008/02/how-c-compiler-decides-garbage-values.html.
First.
Yes, your code is correct (add return 0 command at the end of your main() function though). However, it needs to be more specific, which leads to your second question.
Second.
C is a language that allows a programmer to do lots of things, but it also requires the programmer to do lots of manual coding.
So, when you declare an array of 7 items, C compiler marks a region of memory to accommodate those items (in this case of the type of float). But it doesn't actually check what that region of memory contains until you explicitly assign the values. In your case the last four values (which you have not assigned yourself) are just garbage left in the region of memory marked for your array.
Once again, compiler does not clear the memory for you when you declare the array, it just marks the region of memory. It's your responsibility to assign default values to the array.
Your possible solution is to manually initialize all of the elements of your array to some default value (for example, a 0), like this:
float array[7] = {0}
I have observed that memory allocated for array seems to be dynamic.
Here is the sample code I found in this tutorial:
#include <stdio.h>
main() {
int LA[] = {1,3,5,7,8};
int item = 10, k = 3, n = 5;
int i = 0, j = n;
printf("The original array elements are :\n");
for(i = 0; i<n; i++) {
printf("LA[%d] = %d \n", i, LA[i]);
}
n = n + 1;
while( j >= k){
LA[j+1] = LA[j];
j = j - 1;
}
LA[k] = item;
printf("The array elements after insertion :\n");
for(i = 0; i<n; i++) {
printf("LA[%d] = %d \n", i, LA[i]);
}
}
and sample output:
The original array elements are :
LA[0]=1
LA[1]=3
LA[2]=5
LA[3]=7
LA[4]=8
The array elements after insertion :
LA[0]=1
LA[1]=3
LA[2]=5
LA[3]=10
LA[4]=7
LA[5]=8
How its working I did not get.
First, a general statement, for an array defined without explicit size and initialized using brace-enclosed initializer, the size will depend o the elements in the initializer list. So, for your array
int LA[] = {1,3,5,7,8};
size will be 5, as you have 5 elements.
C uses 0-based array indexing, so the valid access will be 0 to 4.
In your code
LA[j+1] = LA[j];
trying to access index 6, (5+1) which is out of bound access. This invokes undefined behavior.
Output of a code having UB cannot be justified in any way.
That said, main() is technically an invalid signature as per latest C standards. You need to use at least int main(void) to make the code conforming for a hosted environment.
The code has a buffer overflow bug! Arrays in C cannot be extended! You need to allocate enough space when you declare/define it.
You can declare additional space by supplying a size in the declaration:
int LA[10] = {1,3,5,7,8};
LA will now have room for 10 elements with index 0 through 9.
If you want more flexibility you should use a pointer and malloc/calloc/realloc to allocate memory.
Note:
There is a second bug in the copying. The loop starts one step too far out.
With j starting at 5 and assigning index j+1 the code assigns LA[6], which is the 7th element. After the insertion there are only 6 elements.
My conclusion from these 2 bugs is that the tutorial was neither written nor reviewed by an experienced C programmer.
To add on to the other answers, C/C++ do not do any bounds checking for arrays.
In this case you have a stack allocated array, so as long as your index does not leave stack space, there will be no "errors" during runtime. However, since you are leaving the bounds of your array, it is possible that you may end up changing the values of other variables that are also allocated in the stack if it's memory location happens to be immediately after the allocated array. This is one of the dangers of buffer overflows and can cause very bad things to happen in more complex programs.
I'm trying to do a convolution algorithm in C but is stacking on the array of convolution.
#include <stdio.h>
#include <math.h>
#include <stddef.h>
#define convtotal 2590
int main(int argc, char *argv[]) {
int m,n,i,j;
double x[convtotal],h[convtotal];
m=sizeof(x)/sizeof(double);
n=sizeof(h)/sizeof(double);
double aux1[convtotal], aux2[convtotal],conv[convtotal][1];
for (i=0;i<n+m;i++){
if (i<n)
aux1[i]=x[i];
else
aux1[i]=0;
if (i<m)
aux2[i]=h[i];
else
aux2[i]=0;
}
for (i=0;(n+m-1);i++){
conv[i][1]=0;
for (j=0;j<m;j++)
if (i-j+1>0)
conv[i][1]=conv[i][1]+(aux1[j]*aux2[i-j+1]);
}
}
Any suggestions for this problem?
Two problems:
for (i=0;(n+m-1);i++)
You're not limiting i to anything, so the loop doesn't exit when you reach the end of your arrays; it just keeps incrementing i until you hit memory you don't own, at which point you get the segfault. Since conv only goes to m or n, I think you meant to write
for (i = 0; i < n; i++)
Secondly, you declared conv as an Nx1 array, meaning the only legal index in the second dimension may be 0, so the lines
conv[i][1] = 0;
and
conv[i][1]=conv[i][1]+(aux1[j]*aux2[i-j+1]);
should be
conv[i][0] = 0;
and
conv[i][0]=conv[i][0]+(aux1[j]*aux2[i-j+1]);
Not sure why you declared an Nx1 array (seems you could have just declared conv with one dimension), but I may be missing something obvious.
Note that your x and h arrays initially contain random values.
for (i=0;(n+m-1);i++){
conv[i][1]=0;
for (j=0;j<m;j++)
if (i-j+1>0)
conv[i][1]=conv[i][1]+(aux1[j]*aux2[i-j+1]);
}
(n+m-1) infinite loop with constant as a stop condition.
Not actually infinite, runs untill it segfaults.
for (i=0;(n+m-1);i++){
Shouldn't it be i < m+ n - 1?