我在C中制作一个大的整数乘数而没有任何其他来源，如GMP。它可以将10,000个数字的整数乘以另一个。我遇到的问题是经过一定数量的乘法（对我来说，经过大约3次大乘法或25次小乘和3次大）我的程序冻结了。它使用大约1.8 GB的RAM然后停止。我不太确定是什么导致这种情况，我知道不同的数字具有相同的结果。 谢谢你的帮助。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX 10001
struct integer {
int* digits;
int size;
};
struct integer* add(struct integer* one, struct integer *two);
void print(struct integer* number);
void print_op(struct integer* op1, struct integer* op2,struct integer* tempwer,char op);
struct integer* convert_integer(char* word);
void free_struct(struct integer* thisint);
struct integer* multiply(struct integer *p, struct integer *q);
int main() {
FILE* ifp = fopen("bigint.txt", "r");
int loop, numcases;
fscanf(ifp, "%d", &numcases);
// Go through each case.
for (loop=1; loop<=numcases; loop++) {
// Read in both operands.
char op1[MAX];
char op2[MAX];
fscanf(ifp, "%s%s", op1, op2);
// Convert and compute.
struct integer* first = convert_integer(op1);
struct integer* second = convert_integer(op2);
struct integer* ans = multiply(first, second);
printf("Problem #%d: ", loop);
print_op(first, second, ans, '*');
printf("\n");
// After we output, we don't need these any more.
free_struct(first);
free_struct(second);
free_struct(ans);
}
return 0;
}
// Pre-conditions: Both one and two are not NULL pointers that point to
// linked lists structures that store non-negative digits
// at each node.
// Post-conditions: A pointer to a linked list will be returned. The
// linked list will store the value of the sum of the
// two integers represented by the linked lists passed
// into the function.
struct integer* add(struct integer* one, struct integer *two) {
struct integer *ans;
int digit1 = 0, digit2 = 0, carry = 0, result, i;
ans = (struct integer *)malloc(sizeof(struct integer));
//allocate space for the larger of the 2 arrays
if(one->size>two->size)
ans->size=one->size;
else
ans->size=two->size;
ans->digits=(int*)(malloc(sizeof(int)*ans->size));
for(i=0;i<ans->size;i++){
// Determine digit to add from first operand.
if (i<one -> size)
digit1 = one -> digits[i];
else
digit1 = 0;
// Determine digit to add from second operand.
if (i<two -> size)
digit2 = two -> digits[i];
else
digit2 = 0;
// Compute correct addition digit and carry.
result = (digit1 + digit2 + carry)%10;
carry = (digit1 + digit2 + carry)/10;
// Store result in the appropriate linked list.
ans -> digits[i] = result;
}//for
// Take care of the most significant digit if there is a carry.
if (carry != 0) {
//copy off the whole array into a new one
//of size+1 and free the old one in case of carry
ans->size+=1;
ans->digits = (int *)realloc(ans->digits, sizeof(int)*ans->size);
ans->digits[ans->size-1] = carry;
}
// Return the ptr. to the added result.
return ans;
}
// Precondition: number points to a not NULL linked list that contains
// only single digits stored at each node.
// Postcondition: The integer represented by the linked list pointed to by
// number will be printed out.
void print(struct integer* number) {
int i;
if (number != NULL) {
// Loop through in backwards order, since number is stored reversed.
for(i=number->size-1;i>=0;i--){
printf("%d",number->digits[i]);
}
}
}
// Precondition: op1 and op2 point to valid linked lists storing integers,
// operator is a valid integer operator, and tempwer is the
// value of applying the first operation to the second.
// Postcondition: The arithmetic operation desired (op1 operator op2) will
// be printed to the screen in a reasonable manner.
void print_op(struct integer* op1, struct integer* op2,struct integer* tempwer,char op) {
print(op1);
printf(" %c ", op);
print(op2);
printf(" = ");
print(tempwer);
}
//Preconditions: the first parameter is a pointer to a
// pointer variable that points to
// struct integer. The function skips leading
// blanks and assumes that no leading zeros are
// entered at the input.
//Postconditions: The function will read the digits of the
// large integer character by character,
// convert them into integers and place them in
// nodes of a linked list. The pointer to the
// head of the list is returned as the value of
// the input parameter.
struct integer* convert_integer(char* word) {
int i;
struct integer *ans=(struct integer *)(malloc(sizeof(struct integer)));
ans->size=0;
if(word==NULL) ans->digits=NULL;
else {
// Allocate space for each of the digits.
ans->size = strlen(word);
ans->digits=(int *)(malloc(sizeof(int)*ans->size));
// Store them in reverse order.
for(i=0;i< ans->size;i++)
ans->digits[ans->size-i-1]=word[i] - '0';
}//if word not NULL
return ans;
}
//Preconditions: p and q are pointers to struct integers.
//Postconditions: A new struct integer is created that
// stores the product of the integers
// pointed to by p and q and a pointer to it
// is returned.
struct integer* multiply(struct integer *p, struct integer *q){
struct integer *temp;
struct integer *ans;
int digit1 = 0, digit2 = 0, carry = 0, index=0, front=0, result, i, j, pos, preSize;
temp = (struct integer *)calloc(sizeof(struct integer),1);
ans = (struct integer *)calloc(sizeof(struct integer),1);
//allocate space for the larger of the 2 arrays
//Which ever array is larger will be the starting size of the tempwer array.
if(q->size>p->size)
temp->size = q->size;
else
temp->size = p->size;
temp->digits=(int*)(calloc(sizeof(int)*temp->size,1));
//use a double for loop, one for the top number and one for bottom.
for(i=0; i<q->size; i++){
//Make the starting size the size of the biggest number.
if(q->size>p->size)
temp->size = q->size;
else
temp->size = p->size;
temp->digits=(int*)(calloc(sizeof(int)*temp->size,1));
if (i < q->size)
digit1 = q->digits[i];
else
digit1 = 0;
//Bottom part of the multiplication.
for(j=0; j<p->size; j++){
// Determine digit to add from first operand.
if (j < p->size)
digit2 = p->digits[j];
else
digit2 = 0;
// Compute correct multiplication digit and carry.
//gives last digit
result = (digit1 * digit2 + carry)%10;
//drops last digit
carry = (digit1 * digit2 + carry)/10;
// Store result in the appropriate linked list.
temp -> digits[j] = result;
}
//Add a zero to the end of the next number (like multiplying by hand).
if(i>0){
temp->size += i;
temp->digits = (int *)(realloc(temp->digits, sizeof(int)*temp->size));
for(j=temp->size; j>0; j--){
if((j-1-i)<0)
break;
//Shift everthing over by 1.
else
temp->digits[j-1]=temp->digits[j-1-i];
}
//Make the new number zero.
for(j=1; j<=i; j++)
temp -> digits[j-1] = 0;
}//if
//If there is a carry insert it in front of the number.
if (carry != 0) {
temp->size += 1;
temp->digits = (int *)(realloc(temp->digits, sizeof(int)*temp->size));
temp->digits[temp->size-1]=0;
//Find the front of the number.
for(j=temp->size; j>0; j--){
if(temp->digits[j-1] != 0){
front = j;
break;
}
}
//Insert it.
if(result == 0)
temp->digits[front+1] = carry;
else
temp->digits[front] = carry;
carry = 0;
}//if
//Delete any leading zeros.
if(temp->size != 1){
for(j=temp->size; j>0; j--){
//Finds where the zeros end.
if(temp->digits[j-1] != 0){
pos =j;
break;
}
//Counts the zeros.
if(temp->digits[j-1] == 0)
index++;
}
//If they are all zeros then make the temp number 0 of size 1.
if(index==temp->size) {
pos=1;
temp->size = temp->size-(temp->size-pos);
temp->digits = (int *)realloc(temp->digits, sizeof(int)*temp->size);
}
//If not then reallocate with the extra zeros removed.
else{
temp->size = temp->size-(temp->size-pos);
temp->digits = (int *)realloc(temp->digits, sizeof(int)*temp->size);
}
index=0;
}//if
//If this is the first time saving the answer the allocate memory for it.
if(i==0){
ans->size = temp->size;
ans->digits=(int*)(calloc(sizeof(int)*temp->size,1));
}
//Add the temp number to the total (ans).
ans = add(temp, ans);
//Clear temp.
for(j=0; j<temp->size; j++)
temp->digits[j]=0;
}//for on top
free_struct(temp);
return ans;
}
// Frees the memory for the struct pointed to by thisint.
void free_struct(struct integer* thisint) {
free(thisint->digits);
free(thisint);
}
答案 0 :(得分：0)
在你的乘法方法的这一部分：
temp->digits=(int*)(calloc(sizeof(int)*temp->size,1));// <-- allocate some
//use a double for loop, one for the top number and one for bottom.
for(i=0; i<q->size; i++){
//Make the starting size the size of the biggest number.
if(q->size>p->size)
temp->size = q->size;
else
temp->size = p->size;
temp->digits=(int*)(calloc(sizeof(int)*temp->size,1)); // <--allocate again
看起来你正在分配内存，而不是匹配除了最后一个在函数结束时被释放的空闲内容。我使用c已经很多年了，所以希望我没有错过任何明显的东西。