我有一个返回数组指针的函数。我正在循环中运行它,并且 free()
似乎给我带来了问题。我不确定在哪里,但似乎在主循环中的某个地方正在使用我试图释放的内存。我在 10.6 中使用 Xcode 3.2.1 |调试| x86_64 版本。
程序将运行一次主循环;第二次遇到 free()
时,它给出以下错误:
malloc: *** error for object 0x100100180: incorrect checksum for freed object -
object was probably modified after being freed.
有人可以指出(没有双关语)我在这里的指针做错了什么吗?
这是程序:
int main(int argc, char **argv) {
int *partition;
int lowerLimit;
int upperLimit;
// snip ... got lowerLimit and upperLimit from console arguments
// this is the 'main loop':
for (int i = lowerLimit; i <= upperLimit; i += 2) {
partition = goldbachPartition(i);
printOutput(partition[0], partition[1], i);
free(partition); // I get problems on the second iteration here
}
return 0;
}
int *goldbachPartition(int x) {
int solved = 0;
int y, z;
int *primes;
int *result;
result = intAlloc(2);
primes = atkinsPrimes(x);
for (int i = intCount(primes)-1; i >= 0; i--) {
y = primes[i];
for (int j = 0; j < y; j++) {
z = primes[j];
if (z + y >= x) {
break;
}
}
if (z + y == x) {
solved = 1;
result[0] = y;
result[1] = z;
break;
} else if (y == z) {
result[0] = 0;
result[1] = 0;
break;
}
}
free(primes);
return result;
}
int *atkinsPrimes(int limit) {
int *primes;
int *initialPrimes;
int *filtered;
int *results;
int counter = 0;
int sqrtLimit;
int xLimit;
int resultsSize;
primes = intAlloc(limit+1);
intFillArray(primes, limit+1, 0);
sqrtLimit = floor(sqrt(limit));
xLimit = floor(sqrt((limit+1) / 2));
// these loops are part of the Atkins Sieve implementation
for (int x = 1; x < xLimit; x++) {
int xx = x*x;
for (int y = 1; y < sqrtLimit; y++) {
int yy = y*y;
int n = 3*xx + yy;
if (n <= limit && n % 12 == 7) {
primes[n] = (primes[n] == 1) ? 0 : 1;
}
n += xx;
if (n <= limit && (n % 12 == 1 || n % 12 == 5)) {
primes[n] = (primes[n] == 1) ? 0 : 1;
}
if (x > y) {
n -= xx + 2*yy;
if (n <= limit && n % 12 == 11) {
primes[n] = (primes[n] == 1) ? 0 : 1;
}
}
}
}
for (int n = 5; n < limit; n++) {
if (primes[n] == 1) {
for (int k = n*n; k < limit; k += n*n) {
primes[k] = 0;
}
}
}
initialPrimes = intAlloc(2);
if (limit >= 2) {
initialPrimes[counter++] = 2;
}
if (limit >= 3) {
initialPrimes[counter++] = 3;
}
filtered = intFilterArrayKeys(primes, limit+1);
results = intMergeArrays(initialPrimes, filtered, counter, trueCount(primes, limit+1));
resultsSize = counter + trueCount(primes, limit+1);
free(primes);
free(initialPrimes);
free(filtered);
results[resultsSize] = 0;
return results;
}
int trueCount(int *subject, int arraySize) {
int count = 0;
for (int i = 0; i < arraySize; i++) {
if (subject[i] == 1) {
count++;
}
}
return count;
}
int intCount(int *subject) {
// warning: expects 0 terminated array.
int count = 0;
while (*subject++ != 0) {
count++;
}
return count;
}
void intFillArray(int *subject, int arraySize, int value) {
for (int i = 0; i < arraySize; i++) {
subject[i] = value;
}
}
int *intFilterArrayKeys(int *subject, int arraySize) {
int *filtered;
int count = 0;
filtered = intAlloc(trueCount(subject, arraySize));
for (int i = 0; i < arraySize; i++) {
if (subject[i] == 1) {
filtered[count++] = i;
}
}
return filtered;
}
int *intMergeArrays(int *subject1, int *subject2, int arraySize1, int arraySize2) {
int *merge;
int count = 0;
merge = intAlloc(arraySize1 + arraySize2);
for (int i = 0; i < arraySize1; i++) {
merge[count++] = subject1[i];
}
for (int i = 0; i < arraySize2; i++) {
merge[count++] = subject2[i];
}
return merge;
}
int *intAlloc(int amount) {
int *ptr;
ptr = (int *)malloc(amount * sizeof(int));
if (ptr == NULL) {
printf("Error: NULL pointer\n");
}
return ptr;
}
void printOutput(int num1, int num2, int rep) {
if (num1 == 0) {
printf("%d: No solution\n", rep);
exit(0);
} else {
printf("%d = %d + %d\n", rep, num1, num2);
}
}
最佳答案
为什么intAlloc
不返回int*?
int *intAlloc(int amount) {
int *ptr;
ptr = (int *)malloc(amount * sizeof(int));
if(ptr == NULL) {
printf("Error: NULL pointer\n");
exit(1);
}
return ptr; //like this
}
编辑(更新后):
在atkinsPrimes()
上,哪里被intAlloc()ed过滤?
int *atkinsPrimes(int limit) {
int *primes;
int *initialPrimes;
int *filtered;
int *results;
int resultsSize;
primes = intAlloc(limit+1);
// ...
initialPrimes = intAlloc(2);
// ...
resultsSize = counter + trueCount(primes, limit+1);
free(primes);
free(initialPrimes);
free(filtered); // Where was it intAlloc()ed?
results[resultsSize] = 0; // make the array 0-terminated to make it easier to work with
return results;
}
编辑(在您的N次更新之后):
这是您的代码的可编译版本。在我的机器上运行很流畅,没有崩溃。使用 g++ 编译(由于 for 语句内的变量声明):
g++(Debian 4.3.2-1.1)4.3.2
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int *goldbachPartition(int x);
int *atkinsPrimes(int limit);
int trueCount(int *subject, int arraySize);
int intCount(int *subject) ;
void intFillArray(int *subject, int arraySize, int value);
int *intFilterArrayKeys(int *subject, int arraySize);
int *intAlloc(int amount);
void printOutput(int num1, int num2, int rep) ;
int *intMergeArrays(int *subject1, int *subject2, int arraySize1, int arraySize2);
int main(int argc, char **argv) {
if (argc < 3) {
printf("Usage: ./program <lower> <upper>\n");
return 0;
}
int *partition;
int lowerLimit = atoi(argv[1]);
int upperLimit = atoi(argv[2]);
// snip ... got lowerLimit and upperLimit from console arguments
// this is the 'main loop':
for (int i = lowerLimit; i <= upperLimit; i += 2) {
partition = goldbachPartition(i);
printOutput(partition[0], partition[1], i);
free(partition); // I get problems on the second iteration here
}
return 0;
}
int *goldbachPartition(int x) {
int solved = 0;
int y, z;
int *primes;
int *result;
result = intAlloc(2);
primes = atkinsPrimes(x);
for (int i = intCount(primes)-1; i >= 0; i--) {
y = primes[i];
for (int j = 0; j < y; j++) {
z = primes[j];
if (z + y >= x) {
break;
}
}
if (z + y == x) {
solved = 1;
result[0] = y;
result[1] = z;
break;
} else if (y == z) {
result[0] = 0;
result[1] = 0;
break;
}
}
free(primes);
return result;
}
int *atkinsPrimes(int limit) {
int *primes;
int *initialPrimes;
int *filtered;
int *results;
int counter = 0;
int sqrtLimit;
int xLimit;
int resultsSize;
primes = intAlloc(limit+1);
intFillArray(primes, limit+1, 0);
sqrtLimit = floor(sqrt(limit));
xLimit = floor(sqrt((limit+1) / 2));
for (int x = 1; x < xLimit; x++) {
int xx = x*x;
for (int y = 1; y < sqrtLimit; y++) {
int yy = y*y;
int n = 3*xx + yy;
if (n <= limit && n % 12 == 7) {
primes[n] = (primes[n] == 1) ? 0 : 1;
}
n += xx;
if (n <= limit && (n % 12 == 1 || n % 12 == 5)) {
primes[n] = (primes[n] == 1) ? 0 : 1;
}
if (x > y) {
n -= xx + 2*yy;
if (n <= limit && n % 12 == 11) {
primes[n] = (primes[n] == 1) ? 0 : 1;
}
}
}
}
for (int n = 5; n < limit; n++) {
if (primes[n] == 1) {
for (int k = n*n; k < limit; k += n*n) {
primes[k] = 0;
}
}
}
initialPrimes = intAlloc(2);
if (limit >= 2) {
initialPrimes[counter++] = 2;
}
if (limit >= 3) {
initialPrimes[counter++] = 3;
}
filtered = intFilterArrayKeys(primes, limit+1);
results = intMergeArrays(initialPrimes, filtered, counter, trueCount(primes, limit+1));
resultsSize = counter + trueCount(primes, limit+1);
free(primes);
free(initialPrimes);
free(filtered);
results[resultsSize] = 0;
return results;
}
int trueCount(int *subject, int arraySize) {
int count = 0;
for (int i = 0; i < arraySize; i++) {
if (subject[i] == 1) {
count++;
}
}
return count;
}
int intCount(int *subject) {
// warning: expects 0 terminated array.
int count = 0;
while (*subject++ != 0) {
count++;
}
return count;
}
void intFillArray(int *subject, int arraySize, int value) {
for (int i = 0; i < arraySize; i++) {
subject[i] = value;
}
}
int *intFilterArrayKeys(int *subject, int arraySize) {
int *filtered;
int count = 0;
filtered = intAlloc(trueCount(subject, arraySize));
for (int i = 0; i < arraySize; i++) {
if (subject[i] == 1) {
filtered[count++] = i;
}
}
return filtered;
}
int *intMergeArrays(int *subject1, int *subject2, int arraySize1, int arraySize2) {
int *merge;
int count = 0;
merge = intAlloc(arraySize1 + arraySize2);
for (int i = 0; i < arraySize1; i++) {
merge[count++] = subject1[i];
}
for (int i = 0; i < arraySize2; i++) {
merge[count++] = subject2[i];
}
return merge;
}
int *intAlloc(int amount) {
int *ptr;
ptr = (int *)malloc(amount * sizeof(int));
if (ptr == NULL) {
printf("Error: NULL pointer\n");
}
return ptr;
}
void printOutput(int num1, int num2, int rep) {
if (num1 == 0) {
printf("%d: No solution\n", rep);
exit(0);
} else {
printf("%d = %d + %d\n", rep, num1, num2);
}
}
由于您仍然省略了一些来源,我只能想象问题隐藏在那里。
编辑:(我的最后更新)
为了帮助您进行调试,您应该将 main() 函数替换为以下函数:
int main(int argc, char **argv)
{
int *primes = NULL;
primes = atkinsPrimes(44); // Evil magic number
free(primes);
return 0;
}
用一个最小的例子来重现你指出的行为比整个事情要好得多。与 atkinsPrimes(44)
一起享受乐趣关于c - 指针和动态内存,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/4351399/