[英]Game of Life Segmentation Fault in C
對於C,Linux等而言,我是新手,我的代碼可以編譯並運行,但是一旦我輸入第一個用戶輸入,就會遇到分段錯誤。 如果有人指出我的代碼出了什么問題,這將非常有幫助,我認為這是在“ calculate()”或“ main()”中,因為我試圖在兩個方面都使用“ malloc()”來分配內存那些地方。
#include <stdio.h>
#include <stdlib.h>
#define LIFE_YES 'X'
#define LIFE_NO 'O'
int HEIGHT, WIDTH;
typedef int **TableType;
void printTable(TableType table) {
int height, width;
for (height = 0; height < HEIGHT; height++) {
for (width = 0; width < WIDTH; width++) {
if (table[height][width] == LIFE_YES) {
printf("X");
}
else {
printf("-");
}
}
printf("\n");
}
printf("\n");
}
void clearTable(TableType table) {
int height, width;
for (height = 0; height < HEIGHT; height++) {
for (width = 0; width < WIDTH; width++) {
table[height][width] = LIFE_NO;
}
}
}
void askUser(TableType tableA) {
int i;
int n;
int height, width;
printf("Enter the amount of initial organisms: ");
scanf("%d", &n);
for (i = 0; i < n; i++) {
printf("Enter dimensions (x y) where organism %d will live: ", i + 1);
scanf("%d %d", &height, &width);
tableA[height][width] = LIFE_YES;
}
printTable(tableA);
printf("Generation 0");
}
int getNeighborValue(TableType table, int row, int col) {
if (row < 0 || row >= HEIGHT || col < 0 || col >= WIDTH || table[row][col] != LIFE_YES ) {
return 0;
}
else {
return 1;
}
}
int getNeighborCount(TableType table, int row, int col) {
int neighbor = 0;
neighbor += getNeighborValue(table, row - 1, col - 1);
neighbor += getNeighborValue(table, row - 1, col);
neighbor += getNeighborValue(table, row - 1, col + 1);
neighbor += getNeighborValue(table, row, col - 1);
neighbor += getNeighborValue(table, row, col + 1);
neighbor += getNeighborValue(table, row + 1, col - 1);
neighbor += getNeighborValue(table, row + 1, col);
neighbor += getNeighborValue(table, row + 1, col + 1);
return neighbor;
}
void calculate(TableType tableA) {
TableType tableB;
int neighbor, height, width, i;
tableB= malloc(HEIGHT * sizeof(int*));
for (i = 0; i < HEIGHT; i++) {
tableB[i] = malloc(WIDTH * sizeof(int));
}
for (height = 0; height < HEIGHT; height++) {
for (width = 0; width < WIDTH; width++) {
neighbor = getNeighborCount(tableA, height, width);
if (neighbor==3) {
tableB[height][width] = LIFE_YES;
}
else if (neighbor == 2 && tableA[height][width] == LIFE_YES) {
tableB[height][width] = LIFE_YES;
}
else {
tableB[height][width] = LIFE_NO;
}
}
}
for (height = 0; height < HEIGHT; height++) {
for (width = 0; width < WIDTH; width++) {
tableA[height][width] = tableB[height][width];
}
}
free(tableB);
}
/* test data
void loadTestData(TableType table) {
table[3][4] = LIFE_YES;
table[3][5] = LIFE_YES;
table[3][6] = LIFE_YES;
table[10][4] = LIFE_YES;
table[10][5] = LIFE_YES;
table[10][6] = LIFE_YES;
table[11][6] = LIFE_YES;
table[12][5] = LIFE_YES;
}
*/
int main(void) {
int i;
char end;
int generation = 0;
printf("Enter the amount of rows and columns you want in the grid: ");
scanf("%i %i\n", &HEIGHT, &WIDTH);
TableType table = malloc(HEIGHT * sizeof(int*));
for (i = 0; i < HEIGHT; i++) {
table[i] = malloc(WIDTH * sizeof(int));
}
clearTable(table);
askUser(table);
/*loadTestData(table);*/
printTable(table);
while (end != 'q') {
calculate(table);
printTable(table);
printf("Generation %d\n", ++generation);
printf("Press q to quit or 1 to continue: ");
scanf(" %c", &end);
}
return 0;
}
在主要
scanf("%i %i\n", &HEIGHT, &WIDTH); ==> scanf("%i %i", &HEIGHT, &WIDTH);
^
因此\\ n您需要再輸入一個。避免輸入
和askUser
細分錯誤
tableA[height][width] = LIFE_YES;
Program received signal SIGSEGV, Segmentation fault.
0x00000000004007c1 in askUser (tableA=0x603010) at seg3.c:49
49 tableA[height][width] = LIFE_YES;
(gdb) bt
#0 0x00000000004007c1 in askUser (tableA=0x603010) at seg3.c:49
#1 0x0000000000400bf3 in main () at seg3.c:142
(gdb)
在這里您沒有分配內存,但是您正在訪問。首先分配內存
出現段錯誤是因為:
for (i = 0; i < n; i++)
{
printf("Enter dimensions (x y) where organism %d will live: ", i + 1);
scanf("%d %d", &height, &width);
tableA[height][width] = LIFE_YES; <--- Here
}
如果用戶指定的維度超出了您最初在main
動態分配的'HEIGHT'和'WEIGHT'維度的范圍,該怎么辦?
當用戶輸入時,您需要檢查坐標是否為負數並且小於上述尺寸,如果不在范圍之內,則拋出錯誤消息並退出。
第二件事:
另外,main中的end
變量未初始化,也許您在if
條件中錯過了end = getchar()
。
如果未初始化,也可能導致段故障。
第三件事:
\\n
是否真的需要scanf
? 如果存在,它將第三個輸入作為下一個scanf
的輸入,並且不會提示您輸入該scanf的輸入。
也就是說,當前第三個輸入將進入“否”。 沒有提示的生物體。
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