storing Strings into Char array in C & segmentation fault

Question

I'm trying to read input & storing strings in a char array. However, a segmentation fault is returned by the compiler. In addition, storing the string does not work and causes the execution file to crash. Here is my code:

#include <stdlib.h>
#include <math.h>

/*scan functie*/

int inputProducts(int *resourceCost, int *profit, char **productName)  {   
    int amount, i;   
    printf("number of products: \n");   
    scanf("%d", amount);   
    for (i = 0; i < amount; i++)    {
         printf("product: \n");     
         scanf("%s", productName[i]);   
         printf("resource cost for %s: \n", productName[i]);        
         scanf("%d", &resourceCost[i]);     
         printf("profit for %s: \n", productName[i]);   
         scanf("%d", &profit[i]);   
     }   
    return amount;  
}

int main(int argc, char *argv[])    {   
     int amount;    
     int resourceCost[100],profit[100];     
     char *productName[100];    
     amount =  inputProducts(resourceCost, profit, productName);    
     return 0;  
}

compilers don't return segfaults. that's a runtime error. but you're almost certainly exceeding the size of your arrays, e.g. trying to use resourceCost[100], which would only support indexes 0->99. — Marc B
– Marc B, Commented Oct 15, 2014 at 20:28
You'd better check return value of scanf always, because "funny" things will happen when parsing fails but you ignore the error. — hyde
– hyde, Commented Oct 15, 2014 at 20:33
A mistake exists in taking the value of amount aswell. I decided to make the program print it, when I type 5 for the value of amount, it prints 2130567168.... — MD Graaf de
– MD Graaf de, Commented Oct 15, 2014 at 20:43

Mahesh · Accepted Answer · 2014-10-15 20:33:33Z

1

 char *productName[100];

productName is an array of pointers and they are not initialised to point any valid memory locations.

scanf("%s", productName[i]);

And taking input here is causing you segmentation fault.

answered Oct 15, 2014 at 20:33

Mahesh

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3 Comments

MD Graaf de Over a year ago

how would I be able to fix this? Will char productname[100] work?

Mahesh Over a year ago

Hint: Store the address returned by malloc at each index before taking input.

MD Graaf de Over a year ago

Thanks for your solution. I will try this instead. The idea of this exercise however is to store a string, instead of a pointer. But when I initialize the char array without the * , the inputProducts line can't pass the char array (error: incompatible pointer type)

michaelwayman · Accepted Answer · 2014-10-15 20:45:49Z

0

scanf("%d", &amount);   
for (i = 0; i < amount; i++)    {
     productName[i] = (char *)malloc(sizeof(char));
     printf("product: \n");     
     scanf("%s", productName[i]);

you are not allocating any space for the strings to be stored in productName.. try adding this line of code inside your for loop. It will make space on the heap for each of your strings

productName[i] = (char *)malloc(sizeof(char));

edited Oct 15, 2014 at 20:45

answered Oct 15, 2014 at 20:38

michaelwayman

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3 Comments

MD Graaf de Over a year ago

I will try that, however , running the program causes it to crash after typing and entering the amount... Is anything wrong there?

michaelwayman Over a year ago

**corrected my answer. I just ran your program with the correction using malloc() and it worked fine.

MD Graaf de Over a year ago

Thankyou. it works for me too now. The only problem is that I'm storing pointer, as where I would like to store actual values..

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