Valgrind：C程序中绝对会丢失内存泄漏

Question

我正在编写一个迷宫问题程序。 我可以成功获得路径的方向。 但是，我肯定会从valgrind报告中迷失方向和出错。 我的代码有什么问题？

以下是我的代码：

002: #include <stdio.h>
003: #include <stdlib.h>
004: #include <string.h>
005: 
006: struct each_path {
007:     int step_nums, coin_nums;
008:     char **step_dir;
009: };
010: 
011: void initial_value();
012: void *readFile(char *fileName);
013: void maze_1Dto2D(char *array);
014: void visit(int, int);
015: void display_direction();
016: struct each_path *epath;
017: int path_assume_num;
018: char *maze1D, **maze2D, **maze2D_tmp;
019: int maze_height, maze_width;
020: int startI = 1, startJ = 1, endI, endJ;  // entrance & exit
021: int path_nums, coin_nums, min_step_num, min_path_num;
022: 
023: 
024: int main(void) {
025:     char *maze_txtfile;
026:     int i, j;
027: 
028:     // the first maze
029:     initial_value();
030:     maze_txtfile = readFile("maze2.txt");
031:     maze2D_tmp = malloc( maze_height * sizeof(*maze2D_tmp));
032:     for ( i = 0; i < maze_height; ++i ) {
033:         maze2D_tmp[i] = malloc(maze_width + 1);
034:         memcpy(maze2D_tmp[i], maze2D[i], maze_width + 1);
035:     }
036:     epath = malloc(path_assume_num * sizeof(*epath));
037:     for ( i = 0; i < path_assume_num; ++i) {
038:         epath[i].step_dir = malloc(10 * sizeof(*epath[i].step_dir));
039:         for ( j = 0; j < 10; ++j)
040:             epath[i].step_dir[j] = malloc(6);
041:     }
042: 
043:     endI = maze_height - 2;
044:     endJ = maze_width - 2;
045:     visit(startI, startJ);
046:     display_direction();
047: 
048:     for ( i = 0; i < path_nums; ++i) {
049:         for ( j = 0; j < epath[i].step_nums; ++j)
050:             free(epath[i].step_dir[j]);
051:         free(epath[i].step_dir);
052:     }
053:     for ( i = 0; i < maze_height; ++i) {
054:         free(maze2D[i]);
055:         free(maze2D_tmp[i]);
056:     }
057:     free(maze1D);
058:     free(maze2D);
059:     free(maze2D_tmp);
060:     free(epath);
061: 
062:     exit(0);
063: }
064: 
065: void initial_value() {
066:     path_assume_num = 1;
067:     maze_height = 0; 
068:     maze_width = 0;
069:     path_nums = 0; 
070:     coin_nums = 0;
071:     min_step_num = 100000;
072: }
073: 
074: 
075: void *readFile(char *fileName) {
076:     FILE *file = fopen(fileName, "r");
077:     size_t maze_unit_num = 0, maze_assume_size = 100;
078:     int maze_unit;
079: 
080:     if (file == NULL)
081:         return NULL; //could not open file
082: 
083:     maze1D = malloc(maze_assume_size);
084: 
085:     while ((maze_unit = fgetc(file)) != EOF)
086:     {
087:         if (maze_unit_num >= maze_assume_size)
088:         {
089:             maze_assume_size *= 2;
090:             maze1D = realloc(maze1D, maze_assume_size);
091:         }
092:         maze1D[maze_unit_num] = (char) maze_unit;
093: 
094:         if (maze1D[maze_unit_num] == '\n')
095:         {
096:             if (maze_height == 0)
097:                 maze_width = maze_unit_num - 1;
098:             maze_height++;
099:         }
100:         maze_unit_num++;
101:     }
102:     maze1D = realloc(maze1D, maze_unit_num + 1);
103:     maze1D[maze_unit_num] = '\0';
104:     maze_1Dto2D(maze1D);
105: }
106: 
107: void maze_1Dto2D(char *array) {
108:     size_t i = 0, j = 0, num = 0;
109:     maze2D = malloc( maze_height * sizeof(*maze2D));
110: 
111:     for ( i = 0; i < maze_height ; ++i)
112:     {
113:         maze2D[i] = malloc(maze_width + 1);
114:         for ( j = 0; j < maze_width + 1; ++j, ++num)
115:         {
116:             maze2D[i][j] = array[num];
117: 
118:             if (array[num] == '\r')
119:                 --j;
120:             else if (array[num] == '\n')
121:                 maze2D[i][j] = '\0';
122:             else
123:                 maze2D[i][j] = array[num];
124:         }
125:     }
126: }
127: 
128: void visit(int i, int j) {
129:     int preI, preJ, curI = 1, curJ = 1;
130:     int step_nums = 0, step_assume_num = 10; // entrance is not included
131:     char dir[6];
132:     int m, n;
133: 
134:     if (maze2D_tmp[i][j] == '2')
135:         coin_nums++;
136:     maze2D_tmp[i][j] = '3';
137:     if (i == endI && j == endJ) {
138:         if (path_nums >= path_assume_num) {
139:             path_assume_num *= 2;
140:             epath = realloc(epath, path_assume_num * sizeof(*epath));
141:             for ( m = path_assume_num / 2; m < path_assume_num; ++m) {
142:                 epath[m].step_dir = malloc(10 * sizeof(*epath[m].step_dir));
143:                 for ( n = 0; n < 10; ++n)
144:                     epath[m].step_dir[n] = malloc(6);
145:             }
146:         }
147:         while (curI != endI || curJ != endJ)
148:         {
149:             if (step_nums >= step_assume_num) {
150:                 step_assume_num *= 2;
151:                 epath[path_nums].step_dir = realloc(epath[path_nums].step_dir, step_assume_num * sizeof(*epath[path_nums].step_dir));
152:                 for ( m = step_assume_num / 2; m < step_assume_num; ++m)
153:                     epath[path_nums].step_dir[m] = malloc(6);
154:             }
155: 
156:             if ( maze2D_tmp[curI][curJ + 1] == '3' && preJ != (curJ + 1) ) {
157:                 preI = curI;
158:                 preJ = curJ;
159:                 curJ++;
160:                 strcpy(dir, "right");
161: 
162:             }
163:             else if ( maze2D_tmp[curI + 1][curJ] == '3' && preI != (curI + 1) ) {
164:                 preI = curI;
165:                 preJ = curJ;
166:                 curI++;
167:                 strcpy(dir, "down");
168:             }
169:             else if ( maze2D_tmp[curI - 1][curJ] == '3' && preI != (curI - 1) ) {
170:                 preI = curI;
171:                 preJ = curJ;
172:                 curI--;
173:                 strcpy(dir, "up");
174:             }
175:             else if ( maze2D_tmp[curI][curJ - 1] == '3' && preJ != (curJ - 1)) {
176:                 preI = curI;
177:                 preJ = curJ;
178:                 curJ--;
179:                 strcpy(dir, "left");
180:             }
181:             strcpy(epath[path_nums].step_dir[step_nums], dir);
182:             step_nums++;
183:         }
184:         epath[path_nums].step_dir = realloc(epath[path_nums].step_dir, step_nums * sizeof(*epath[path_nums].step_dir));
185:         epath[path_nums].step_nums = step_nums;
186:         epath[path_nums].coin_nums = coin_nums;
187:         path_nums++;
188:         if (step_nums < min_step_num)
189:         {
190:             min_step_num = step_nums;
191:             min_path_num = path_nums;
192:         }
193:     }
194: 
195:     if (maze2D_tmp[i][j + 1] == '1' || maze2D_tmp[i][j + 1] == '2') visit(i, j + 1);
196:     if (maze2D_tmp[i + 1][j] == '1' || maze2D_tmp[i + 1][j] == '2') visit(i + 1, j);
197:     if (maze2D_tmp[i][j - 1] == '1' || maze2D_tmp[i][j - 1] == '2') visit(i, j - 1);
198:     if (maze2D_tmp[i - 1][j] == '1' || maze2D_tmp[i - 1][j] == '2') visit(i - 1, j);
199: 
200:     if (maze2D[i][j] == '2')
201:     {
202:         maze2D_tmp[i][j] = '2';
203:         coin_nums--;
204:     }
205:     else
206:         maze2D_tmp[i][j] = '1';
207: }
208: 
209: void display_direction() {
210:     int i;
211:     for ( i = 0; i < min_step_num; ++i)
212:         printf("%s\n", epath[min_path_num - 1].step_dir[i]);
213:     printf("coin numbers:%d\n", epath[min_path_num - 1].coin_nums);
214: }

Valgrind报告：

我从泄漏摘要中得到了“绝对丢失：318个块中的1,908字节”，从错误摘要中得到了来自4个上下文的4个错误（被抑制：从0开始为0）。 是什么导致这些错误消息？

 ==3410== HEAP SUMMARY: ==3410== in use at exit: 2,460 bytes in 319 blocks ==3410== total heap usage: 742 allocs, 423 frees, 37,249 bytes allocated ==3410== ==3410== 954 bytes in 159 blocks are definitely lost in loss record 2 of 3 ==3410== at 0x4C2DB8F: malloc (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so) ==3410== by 0x401007: visit (4thesecondmaze.c:153) ==3410== by 0x401352: visit (4thesecondmaze.c:196) ==3410== by 0x401352: visit (4thesecondmaze.c:196) ==3410== by 0x401352: visit (4thesecondmaze.c:196) ==3410== by 0x401352: visit (4thesecondmaze.c:196) ==3410== by 0x401352: visit (4thesecondmaze.c:196) ==3410== by 0x401352: visit (4thesecondmaze.c:196) ==3410== by 0x401352: visit (4thesecondmaze.c:196) ==3410== by 0x4012EC: visit (4thesecondmaze.c:195) ==3410== by 0x4012EC: visit (4thesecondmaze.c:195) ==3410== by 0x4012EC: visit (4thesecondmaze.c:195) ==3410== ==3410== 954 bytes in 159 blocks are definitely lost in loss record 3 of 3 ==3410== at 0x4C2DB8F: malloc (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so) ==3410== by 0x401007: visit (4thesecondmaze.c:153) ==3410== by 0x4012EC: visit (4thesecondmaze.c:195) ==3410== by 0x401352: visit (4thesecondmaze.c:196) ==3410== by 0x4013BA: visit (4thesecondmaze.c:197) ==3410== by 0x401352: visit (4thesecondmaze.c:196) ==3410== by 0x401352: visit (4thesecondmaze.c:196) ==3410== by 0x401352: visit (4thesecondmaze.c:196) ==3410== by 0x401352: visit (4thesecondmaze.c:196) ==3410== by 0x401352: visit (4thesecondmaze.c:196) ==3410== by 0x401352: visit (4thesecondmaze.c:196) ==3410== by 0x4012EC: visit (4thesecondmaze.c:195) ==3410== ==3410== LEAK SUMMARY: ==3410== definitely lost: 1,908 bytes in 318 blocks ==3410== indirectly lost: 0 bytes in 0 blocks ==3410== possibly lost: 0 bytes in 0 blocks ==3410== still reachable: 552 bytes in 1 blocks ==3410== suppressed: 0 bytes in 0 blocks ==3410== Reachable blocks (those to which a pointer was found) are not shown. ==3410== ERROR SUMMARY: 4 errors from 4 contexts (suppressed: 0 from 0) 

Answer 1

在第153行，您正在将一堆内存块分配到step_dir ，最后得到step_assume_num分配的块。

稍后，在第184行，您调用realloc以将为step_dir分配的内存减少到实际使用的数量step_num块。 分配但未使用的块（从step_num + 1到step_assume_num ）尚未释放。 您需要在减小阵列大小之前释放这些块。