我对我遇到的错误感到非常绝望。
在大学的C编程课上,我必须实现一个解析器来解析GML(图形建模语言)输入流。
成功后,解析器将一个抽象数据类型作为邻接矩阵返回给调用者,以表示该图形。
好吧,解析器运行得很完美,如果不是最近几天一直困扰着我,就不会有问题了。在解析器中,有一个函数调用,它反过来调用malloc。当扫描程序逐个符号传递给解析器时,malloc经常被调用。但是,在离开扫描程序例程之前,始终通过调用free()释放malloc分配的内存块。
但是,在解析器深入的一个致命函数调用中,又调用一个使用malloc来保留12个字节内存(三个整数属性)以保存结构体的函数。该结构体用于存储图中单个边的信息(源节点、目标节点、权重)。
这个调用被执行了两次。第一次,一切正常。然后,根据gml语法可以出现1到n条边,代码进入while循环,在其中将相同的指针分配给指向新Edge Struct的指针,只要在输入流中找到边。循环中的Edge识别例程的第一个调用,总共是第二个调用(第一个调用发生在进入循环之前,请参见m.a.),不断失败,malloc返回NULL。
我真的不知道为什么。
这不是关于内存短缺问题,因为当我在该程序的main()函数中malloc 1000字节时,只是为了好玩,它可以正常工作。
我使用Code::Blocks和DevCPP作为IDE。在两者中,程序都遇到了同样的问题。
这是我的主解析例程:
DirectedGraph Graph(char* sourceString, int*currentPosition){
int sym;
int restartPosition = 0;
int* backupPosition;
char* backupString;
int nodeCount = 0;
int currentSrc = -1;
int currentTgt = -1;
int currentWgt = -1;
EdgeDescription e;
DirectedGraph correctMatrix;
MatrixStruct* errorMatrix = NULL;
/*begin parsing*/
bool isGraphHeader = GraphHdr(sourceString, currentPosition);
if(isGraphHeader == true){
bool isNode = Node(sourceString, currentPosition);
if(isNode == true){
while(isNode == true){
nodeCount++;
restartPosition = *currentPosition;
isNode = Node(sourceString, currentPosition);
}
*currentPosition = restartPosition;
/*now get edge information (from-to-weight)*/
/*as we have already read the next symbol, we have to reset*/
/*our read position by one symbol backwards*/
e = Edge(sourceString, &restartPosition); /*<======== HERE I CALL THE FATAL ROUTINE FOR THE FIRST TIME - EVERYTHING´s JUST FINE, PROGRAM PROCEEDS*/
restartPosition = 0;
/*just for clearer coding in if statement*/
currentSrc = e->source;
currentTgt = e->target;
currentWgt = e->weight;
destroyEdge(e);
if(currentSrc != -1 && currentTgt != -1 && currentWgt != -1){
/*initialize matrix with counted number of nodes*/
correctMatrix = CreateNewGraph(nodeCount);
/*the edge is inserted only when it lies within the boundaries*/
/*of our graph. but we do not interrupt the whole processing, we just skip it.*/
while(currentSrc != -1 && currentTgt != -1 && currentWgt != -1){
if(currentSrc <= nodeCount && currentTgt <= nodeCount){
InsertEdge(correctMatrix, currentSrc, currentTgt, currentWgt);
restartPosition = *currentPosition;
}
e = Edge(sourceString, currentPosition); /* <============== THIS IS THE CALL THAT FAILS*/
currentSrc = e->source;
currentTgt = e->target;
currentWgt = e->weight;
}
/*as we have read over the next symbol in the loop, reset the position to read*/
*currentPosition = *currentPosition - 1;
sym = GetNextSymbol(sourceString,currentPosition);
if(sym == rightBrace){
sym = GetNextSymbol(sourceString, currentPosition);
if(sym == eot){
return correctMatrix;
}
else{
return errorMatrix;
}
}
else{
return errorMatrix;
}
}
else{
return errorMatrix;
}
}
else{
return errorMatrix;
}
}
else{
return errorMatrix;
}
在这里是 GetNextSymbol(即 scanner,用于向解析器提供符号):
/**
* DOCUMENTATION
* ============================
* This is the main scanning function
* which is used by the parser to recognize
* terminal symbols and valid literals.
*
* RETURNS: the enum code for the recognized symbol.
* or an error code, when invalid symbol encountered.
*/
int GetNextSymbol(char* sourceString, int* currentPosition){
int symbolCode;
int loopCounter = 0;
char* currentIdentifier = (char*)malloc(10);
char* currentNumber = (char*)malloc(10);
int identifierPosition = 0;
int numberPos = 0;
int numericVal = 0;
char currentChar;
currentChar = getNextChar(sourceString, currentPosition);
/*skip all blanks, empty chars,
linefeeds, carriage returns*/
while(currentChar == ' '
|| currentChar == 11
|| currentChar == 10
|| currentChar == 13
|| currentChar == '\t')
{
currentChar = getNextChar(sourceString, currentPosition);
}
/*=====================================*/
/*Section 1: scan for terminal symbols */
/*====================================*/
if(currentChar == '['){
symbolCode = leftBrace;
}
else if(currentChar == ']'){
symbolCode = rightBrace;
}
/*=====================================*/
/*Section 2: scan for valid literals */
/*====================================*/
else if(isdigit(currentChar)){
/*here we calculate the numeric value of a number expression*/
/*when calculated, we assign the numeric value to the symCode variable*/
/*this works out because the values for a real symbol are always negative*/
symbolCode = digit;
while(isdigit(currentChar)){
currentNumber[numberPos] = currentChar;
currentChar = getNextChar(sourceString, currentPosition);
loopCounter++;
numberPos++;
}
currentNumber[numberPos] = '\0';
numericVal = atoi(currentNumber);
symbolCode = numericVal;
/*when identifier or braces follow number without space: reset currentPos*/
/*to the position of the previous char*/
if(isalpha(currentChar)){
*currentPosition = *currentPosition - loopCounter;
}
else if(currentChar == ']'){
*currentPosition = *currentPosition - loopCounter;
}
else if(currentChar == '['){
*currentPosition = *currentPosition - loopCounter;
}
}
else if(isalpha(currentChar)){
while(isalpha(currentChar)){
currentIdentifier[identifierPosition] = currentChar;
currentChar = getNextChar(sourceString, currentPosition);
loopCounter++;
identifierPosition++;
}
/*check wether we have found a valid identifying label*/
/*and deallocate the reserved mem space*/
currentIdentifier[identifierPosition] = '\0';
symbolCode = recognizeIdentifier(currentIdentifier);
/*when number or braces follow identifier without space: reset currentPos*/
/*to the position of the previous char*/
if(isdigit(currentChar)){
*currentPosition = *currentPosition - 1;
}
else if(currentChar == ']'){
*currentPosition = *currentPosition - 1;
}
else if(currentChar == '['){
*currentPosition = *currentPosition - 1;
}
}
else if(currentChar=='\0'){
symbolCode = eot;
}
/*neither terminal symbol nor end of text found on current position --> illegal symbol*/
else{
symbolCode = error;
}
free(currentIdentifier);
free(currentNumber);
return symbolCode;
}
现在是“Edge”识别例程中致命调用。首先,结构体的头部。
#ifndef GML_EDGE_STRUCT_H_INCLUDED
#define GML_EDGE_STRUCT_H_INCLUDED
typedef struct EdgeStruct* EdgeObj;
typedef struct EdgeStruct {
int source;
int target;
int weight;
} EdgeStruct;
typedef EdgeObj EdgeDescription;
EdgeDescription createNewEdge(int src, int tgt, int wgt);
void destroyEdge(EdgeObj);
#endif // GML_EDGE_STRUCT_H_INCLUDED
实现方式
#include "GML_EDGE_STRUCT.h"
#include <stdio.h>
#include <stdlib.h>
EdgeDescription createNewEdge(int source, int target, int weight){
EdgeDescription e;
int bytesRequested = sizeof(EdgeStruct);
e = malloc(bytesRequested);
e->source = source;
e->target = target;
e->weight = weight;
return e;
}
我知道,那是相当多的代码 ;) 只是为了表明,一切可以释放的东西,我都已经释放了。
我在过去两天里谷歌了我的问题,当然也在这里的stack overflow上找到了数百个网站、帖子等等,都涉及到malloc返回null。它们基本上都说了同样的话:内存不足(这让人有点难以置信),或堆被分段,因此没有足够大小的内存块可用。
但是:我所请求的仅仅是12(十二)字节来存储三个int属性。这似乎太多了。
我是否超出了我不知道的某些内部限制?
非常感激您的帮助。
提前表示感谢 Roland
编辑于2012年11月24日:
谢谢你们的回答。 但是。问题必须是更基本的性质。
因为:当我测试程序的其他部分(文件I/O等)时,它们比解析器复杂得多,只需从main()调用一次,我也无法分配内存。我读取的文件大约有140字节。即使我将I/O部分与所有其他部分隔离开来进行测试,即使我将它们外包给另一个项目,我也无法从系统中获得内存。我已经重新启动了计算机,一切都是绝对的,没有任何改变。
有什么想法吗? 与此同时,我在这个项目上花费了太多时间,其中大部分时间都是用来跟踪那些可恶的内存错误…… :-(((