Question about compile error of atoi in c language

Question

The problem is that the line with "atoi" can not be compiled. If I try to debug my code, this statement comes out "Unhandled exception at 0x7C17F2F6 (ucrtbased.dll) in A5(Dynamic Memory allocation).exe: An invalid parameter was passed to a function that considers invalid parameters fatal."

I think I didn't missed any single thing for the atoi part. I am using visual studio, and would that be the problem? Shall I change the program??

This is the code:

#include <stdio.h>
#include <stdlib.h>

double arith_seq(double*, int, int);

int main(int argc, char* argv[])
{
double* in;
int num;
num = atoi(argv[1]);

in = (double*)malloc((num+1) * sizeof(double));

if (in == NULL)
{
    num = 1;

    arith_seq(&in, num, 0.1);
}
//if there is no input

else
{
    in[0] = 0;

    arith_seq(&in, num, 0.1);
}
//in normal case

printf("My last array element is : %f", in[num - 1]);
//print the last element of the array

free(in);
return 0;
}

double arith_seq(double* parr, int size, int com)
{
for (int i = 1; i < size; i++)
{
    parr[i] += parr[i - 1] + com;
}
return *parr;
}

I think `atoi` in the code must have compiled. Always check `argc` before accessing `argv`. — Weather Vane, May 04 '21 at 18:25
'arith_seq' expects a pointer to a 'double' as first argument, you are passing a pointer to a pointer to a 'double'. Instead of passing '&in', pass 'in' — lulle2007200, May 04 '21 at 18:25
First of all, never use any element of``argv` before checking `argc` to make sure the element in `argv` is valid. — Some programmer dude, May 04 '21 at 18:26
⟼Remember, it's always important, *especially* when learning and asking questions on Stack Overflow, to keep your code as organized as possible. [Consistent indentation](https://en.wikipedia.org/wiki/Indentation_style) helps communicate structure and, importantly, intent, which helps us navigate quickly to the root of the problem without spending a lot of time trying to decode what's going on. — tadman, May 04 '21 at 18:26
Secondly, please learn how to use an actual debugger to catch crashes like these. — Some programmer dude, May 04 '21 at 18:26
@WeatherVane It's not required, but it is a bad idea to access 'argv' without checking 'argc'. The moment you dont call your program without an argument, you get undefined behaviour when accessing 'argv[1]'. Worst case you dont even get any error, best case you get an access violation and know you messed up. — lulle2007200, May 04 '21 at 18:27
@lulle please address your remark to OP. I think I already mentioned that `argc` should be checked, and I do understand why. — Weather Vane, May 04 '21 at 18:28
@WeatherVane How can I check argc whether it exists or not, even if I don't use it in my code? — hong, May 04 '21 at 18:39
With `if(argc > 1) { num = atoi(argv[1]); } else { /* issue error message */ }`. That is *why* `argc` is given to you. A related question is [What are the arguments to main() for?](https://stackoverflow.com/questions/3734111/what-are-the-arguments-to-main-for) — Weather Vane, May 04 '21 at 18:41

lulle2007200 · Accepted Answer · 2021-05-04T19:07:06.583

There are multiple things wrong.

You access argv[1] without checking argc first to know if there even are any arguments stored in argv
double arith_seq(double* parr, int size, int com) expects a double pointer as first argument. You are passing a pointer to a double pointer at multiple places (e.g. arith_seq(&in, num, 0.1) in has type double*, you are passing the address of that)
double arith_seq(double* parr, int size, int com) expects an int as third argument, you are passing a double at multiple places (e.g. arith_seq(in, num, 0.1), 0.1 is not an int). I don't think you want to do that.
malloc expects a size_t argument, but you are passing (num + 1) * sizeof(double) . What if (num + 1) is negative? That will lead to some "interesting" behaviour (Think about what unsigned value -1 represents for example).
You check if malloc returned a NULL pointer (in == NULL), but still go ahead and call arith_seq, which accesses elements of in. You are not allowed to dereference a NULL pointer.
You refer to in[num-1] as the last array element, but actually in[num] is the last elelement. Remember, you allocated an array of num+1 elelements.
In arith_seq you do parr[i] += parr[i-1] + com, which is equal to parr[i] = parr[i] + parr[i-1] + com. But parr[i] has not been initialized anywhere in your code and contains garbage data. This garbage data propagates through the entire array in your loop.

I recommend to start over with that code. I am not exactly sure what you intend to do with the code, so i can't fully fix it (this fixes 1 and 2, for the rest i don't even know what your original intention was), but this atleast doesnt crash:

#include <stdio.h>
#include <stdlib.h>

double arith_seq(double*, int, int);

int main(int argc, char* argv[])
{
    double* in;
    int num;
    if(argc <= 1){
        return(1);
    }
    num = atoi(argv[1]);
    printf("%d\n", num);
    in = (double*)malloc((num+1) * sizeof(double));
    if (in == NULL)
    {
        return(1);
    }
    in[0] = 0.0f;
    arith_seq(in, num, 0.1);

    printf("My last array element is : %f", in[num]);

    free(in);
    return(0);
}

double arith_seq(double* parr, int size, int com)
{
    for (int i = 1; i < size; i++)
    {
        parr[i] += parr[i-1] + com;
    }
    return *parr;
}

Could simplify `in = (double*)malloc((num+1) * sizeof(double));` with `in = malloc(sizeof *in * (num+1u));` — chux - Reinstate Monica, May 04 '21 at 19:18

Question about compile error of atoi in c language

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