How to extract numbers from a string and add them all up c

Question

In my programming, the user enters a string containing numbers, and the system needs to dynamically allocate an array of integers to receive these numbers, converting the character numbers in the string to the integer numbers in the int array one at a time. And then you add up these numbers. That is my code.

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

int main(void) {
    char integral[10]; // create an char array to store at most 9 letters
    printf("Enter a string that include number: ");
    scanf("%9s", integral); // record the string
    int length = strlen(integral); // get the whole length of array after users finished 
    int *array; // create an integral pointer
    array = malloc(sizeof(int) * (length + 1)); // Dynamically allocate an array size that can hold exactly all the numbers

    strcpy(array, integral); // copy the number from array to integral
    int sum = 0;
    for (int i = 0; i < length; i++) {
        sum += array[i]; // get the sum
    }
    free(array);
    printf("The sum is : %d", sum); // print the result
    return 0;
}

But there are some errors happened after complied:

1. warning: passing argument 1 of 'strcpy' from incompatible pointer type on line 13.
2. note: expected 'char * restrict' but argument is of type 'int *'.

How to modify my code?

Answer 1

What you are trying to do can NOT work. You try to use the strcpy(restrict char*, restrict char*) , even if it can only take two char pointer which do not point to the same address. However, you try to pass an int pointer to the function.

The reason for your code being utterly erroneous is quite simple : a char is encoded on 1 byte, whereas an int is encoded on 2 or 4 bytes (depending on your architecture).

Moreover, even if by whatever shenanigan your compiler accepted to copy as you wish, there would be no conversion. Instead, you would just have the ascii value. For example, '1' would give 49.

You must not use the strcpy function and you must convert your characters.

So assuming you have to use a dynamic array and assuming the string given by the user contains only digits , you should write :

for (int i = 0; i < length; ++i) {
    array[i] = integral[i] - '0';
}

Then you can compute your sum

Answer 2

As commented by others, you do not have to dynamically allocate an array of integers just to calculate the sum of them (unless required by your instructor). Would you please try:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <ctype.h>
#define LEN 10

int main(void) {
    int sum = 0;
    char integral[LEN];                         // char array to store at most LEN-1 letters
    char *p = integral;                         // a pointer within "integral"
    char *err;                                  // remaining string after strtol

    printf("Enter a string that include number: ");
    fgets(integral, LEN, stdin);

    while (1) {
        sum += (int)strtol(p, &err, 10);
        for (p = err; *p && !isdigit(*p); p++); // increment pointer to next digit
        if (! *p) break;                        // end of string
    }

    printf("The sum is : %d\n",sum);            // print the result
    return 0;
}

If you are forced to use an array of integers, here is an alternative with an array:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <ctype.h>
#define LEN 10

int main(void) {
    int sum = 0;
    char integral[LEN];                         // char array to store at most LEN-1 letters
    char *p = integral;                         // a pointer within "integral"
    char *err;                                  // remaining string after strtol
    int *ary = NULL;                            // array of integers
    int i;
    int n = 0;                                  // count of integers

    printf("Enter a string that include number: ");
    fgets(integral, LEN, stdin);

    for (; *p && !isdigit(*p); p++);            // increment pointer to next digit
    while (1) {
        ary = realloc(ary, sizeof(int) * (n + 1));
        ary[n++] = (int)strtol(p, &err, 10);
        for (p = err; *p && !isdigit(*p); p++); // increment pointer to next digit
        if (! *p) break;                        // end of string
    }

    for (i = 0; i < n; i++) {
        sum += ary[i];
    }
    printf("The sum is : %d\n",sum);            // print the result

    free(ary);
    return 0;
}