如何乘以 2 个大数字作为字符串（以前的幻想添加了 2 个字符串）？ C

Question

verylong multiply_verylong(verylong vl1, verylong vl2)'verylong defines as typedef char* verylong'

{

size_t maxln, minln;
int carry=0, placeSaver=1, i, k ,sum=0,dif,newln,newln2,ln1,ln2;
verylong newNum , maxVl,minVl,tempvl,addvl;
ln1 = strlen(vl1); // 'length of first str'
ln2 = strlen(vl2);'length of second str'

if (ln1 >= ln2)
{


    maxln = ln1;
    minln = ln2;
    maxVl = (verylong)calloc(maxln + 1, sizeof(char));
    assert(maxVl);
    minVl = (verylong)calloc(minln + 1, sizeof(char));
    assert(minVl);
    strcpy(maxVl, vl1); 
    strcpy(minVl, vl2);
    dif = maxln - minln;
}


else 'stops debuging here'
{

    maxln = ln2;
    minln = ln1;
    maxVl = (verylong)calloc(maxln + 1, sizeof(char));
    assert(maxVl);
    minVl = (verylong)calloc(minln + 1, sizeof(char));
    assert(minVl);
    strcpy(maxVl, vl2);
    strcpy(minVl, vl1);
    dif = maxln - minln;
}

newln = 2 * maxln + 1; 'maximum length of new  required string'
newln2 = newln - 1; 'the index of the new string'

newNum = (verylong)calloc(newln,sizeof(char)); 
addvl = (verylong)calloc(newln, sizeof(char));
tempvl = (verylong)calloc(newln, sizeof(char));
for (i = minln - 1; i >= 0; i--) ' elementry school multiplication'
{


    for (k = maxln - 1; k >= 0; k--)
    {

        sum = ((minVl[i] - '0')*(maxVl[k] - '0')*placeSaver)+carry;
        if (sum >= 10)
            carry = sum / 10;
        if (k == 0)
            newNum[newln2] = '0' + sum;
        else
            newNum[newln2] = '0' + sum%10;
    newln2--;

    }


    placeSaver*=10; 
    addvl=add_verylong(newNum,tempvl);'sending the 2 strings to a previous function that adds 2 strings'
    strcpy(tempvl, addvl);


}


return addvl;

}

void main() {

char vl1[80], vl2[80];
printf("enter  large number\n"); 
gets(vl1);
printf("enter  large number\n");
gets(vl2);
verylong res = multiply_verylong(vl1, vl2);'saves the string  '
printf("%s", res);
free(res);

}

I tried to multiply the first digit of the first number from right with all of the digits from the second number moving forward to the second digit of the first number and then to multiply the placesaver by 10 .

***

• the problem is that the code outputs usually nothing and sometimes just inccorect result

---

***

Answer 1

Your approach is basically right, but there are some mistakes in the implementation.

• You allocate space for addvl , but overwrite that pointer later with the return value of add_verylong() . This gives a memory leak. Further memory leaks are produced by not freeing tempvl , maxvl , minvl and newNum .
• You forgot to initialize tempvl to the number "0" .
• We cannot shift the intermediate product by multiplying each digit with the power of ten placeSaver . What we can do is place 0 digits to the right of the product newNum .
• You failed to correctly set carry in each loop cycle.
• You missed that even the leftmost digits can produce a carry .

This code has the mentioned mistakes corrected:

    newNum = calloc(newln, sizeof(char)); 
    // do not allocate space for addvl - it would be lost
    tempvl = malloc(newln);
    strcpy(tempvl, "0");    // don't forget to initialize tempvl to "0"
    for (i = minln - 1; i >= 0; i--)    // elementry school multiplication
    {
        newln2 = newln - 1; // the index of the new string
        for (carry = 0, k = maxln - 1; k >= 0; k--) // clear carry before each loop
        {
            sum = (minVl[i] - '0')*(maxVl[k] - '0') + carry;
            carry = sum/10;
            newNum[--newln2] = '0' + sum%10;
        }
        if (carry) newNum[--newln2] = '0' + carry;
        addvl = add_verylong(newNum+newln2, tempvl);    // number starts at +newln2
        free(tempvl);                                   // free obsolete number
        tempvl = addvl;                                 // rather than strcpy()
        newNum[--newln-1] = '0';                        // instead of placeSaver*=10
    }
    free(maxVl), free(minVl), free(newNum);
    return addvl;

It works for the example multiply_verylong("23345", "34565") , but you should do further tests.