signal: segmentation fault (core dumped) error

Question

Wrote a program that multiply two arrays like this:

uv = u1v1 + u2v2 + u3v3 + ... un*vn

after getting from the user both of the arrays, i get a "signal: segmentation fault (core dumped)" error.

here's the code:

#include <stdio.h>


int scalar_product(int vectorU[], int vectorV[], int vectorLength) {
  int i, sum = 0;
  for (i = 0; i < vectorLength; i++)
    sum += (vectorU[i] * vectorV[i]);
  return sum;
}

void userInterface() {
  int vectorLength = 0, i;
  printf("Please enter the length of the vectors: ");
  scanf("%d", &vectorLength);
  int vectorU[vectorLength], vectorV[vectorLength];
  printf("\nVector U:");
  for (i = 0; i < vectorLength; i++) {
    printf("\n%d) ", (i + 1));
    scanf("%d", &vectorU[i]);
  }
  printf("\nVector V:");
  for (i = 0; i < vectorLength; i++) {
    printf("\n%d) ", (i + 1));
    scanf("%d", &vectorV[i]);
  }
  printf(scalar_product(vectorU, vectorV, vectorLength));
}

main(void) {
  userInterface();
}

Always enable and heed your compiler's warnings. (e.g. -Wall -Wextra -pedantic with gcc) The problem would have been identified. — ikegami
– ikegami, Commented Mar 31, 2022 at 19:48
As an aside, main(void) { ... } should be int main(void) { ... } — ikegami
– ikegami, Commented Mar 31, 2022 at 19:49
Segmentation fault usually means you have tried to access something that is not a valid pointer reference as a pointer. core dumped means that you have something on disk to analyze. You may want to learn to do this using gdb - it will help you a lot. — Thorbjørn Ravn Andersen
– Thorbjørn Ravn Andersen, Commented Oct 8, 2024 at 14:22

Vlad from Moscow · Accepted Answer · 2022-03-31 23:36:10Z

0

This call of printf is incorrect

printf(scalar_product(vectorU, vectorV, vectorLength));

You need to write at least

printf( "%d\n", scalar_product(vectorU, vectorV, vectorLength));

Also it would be better to declare and define the function like

long long int scalar_product( const int vectorU[], const int vectorV[], int vectorLength) {
  long long int sum = 0;
  for ( int i = 0; i < vectorLength; i++)
    sum += ( long long int )vectorU[i] * vectorV[i];
  return sum;
}

To output the result you need to use the format string "%lld\n"..

The type long long int is used to avoid an overflow.

Another way is to declare the function return type as double.

Also you forgot to specify the return type int of the function main.

edited Mar 31, 2022 at 23:36

answered Mar 31, 2022 at 19:46

Vlad from Moscow

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Comments

John Bollinger · Accepted Answer · 2022-03-31 19:52:24Z

Here:

  printf(scalar_product(vectorU, vectorV, vectorLength));

... you have failed to specify a format for printf. As a result, it attempts to interpret the result of scalar_product() as a pointer to the format string. Undefined behavior results.

If your compiler is not emitting a warning about that then you should either learn how to turn up the warning level so that it does, or else get a better compiler. If your compiler is emitting a warning about it, then take this as a lesson that it is not safe to ignore compiler warnings.

Probably you wanted somethign more like this:

  printf("%d\n", scalar_product(vectorU, vectorV, vectorLength));

As a minor additional issue, you have forgotten the return type for main(). Your compiler is probably treating it as returning int, which turns out to be the right thing to do, but that doesn't make your code correct. You want:

int main(void) {
    // ...

With those two changes, your program compiles for me without any diagnostics, and runs without error, producing the result I expect.

At least, for small vector lengths. If you try really large vectors then you might exceed the available space for allocating your vectors on the stack.

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