分段错误错误：同时减去两个数组

Question

I cannot subtract two arrays which are declared using malloc.what is the problem?

#include<stdio.h>
#include<conio.h>
void main()
{
float *a,*b,*c;
int i;
a=(float*)malloc(8*sizeof(float));
b=(float*)malloc(8*sizeof(float));
c=(float*)malloc(8*sizeof(float));
a[1]=8.0;a[2]=9.0;a[3]=1.0;a[4]=2.0;a[5]=0.0;a[6]=8.0;a[7]=18.0;a[8]=2.0;
b[1]=10.0;b[2]=1.0;b[3]=13.0;b[4]=0.0;b[5]=100.0;b[6]=1.0;b[7]=20.0;b[8]=0.0;
for(i=1;i<=8;i++)
{
c[i]=a[i]-b[i];
printf("%d",c[i];// IAM GETTING ERROR HERE
}
getch();
}

Answer 1

a = (float*)malloc(8*sizeof(float));

This will allocate 8 continuous chunks , each of size equal to size of float and returns pointer to the first chunk and the value at each chunk can be accessed as

a[0], a[1] ,..., a[7] or *(a), *(a+1), ..., *(a+7)

In your code, I can see a[8], b[8], c[8], modify your code to go from 0 to 7 rather than from 1 to 8

Answer 2

Gautam seems to have already answered your question in the comment, and I'd like to make it clearer.

Segmentation Fault is almost always due to inappropriate memory usage, and in your case the culprit is c[8]. Remember that in languages like C, you count from 0, that is, if the length of an array is 8, its index should be 0, 1, ... 7. In fact, the indices are more like offsets; the first element doesn't need any offset and hence with index 0, the second element need to have one offset (the size of an array element) added, the third needs two, the fourth needs three, et cetera.

For malloc(): every chunk of memory allocated should be freed eventually to avoid memory leak. In your situation, why not just declare three arrays with length eight? That'll spare you from worrying about these.

Format String: "%d" is used to print out decimals, while "%f" is for floats. If you wanna see floating point numbers on the screen "%f" is what you should go for. (also the right parenthesis is missing in your printf(). I assume it's just a typo)