好的,我知道这里有另外一个关于“exc_bad_access”的问题,但这似乎与Objective-C和iPhone开发有关,而我的只是常规C.我是C的新手,我几乎完成了我的第一次程序,直到出现此错误。我一直试图弄明白这一天,我疯了。任何帮助表示赞赏。 越野车功能:
void edit (int i){
char* z;
char* y;
char compare1[] = "on bobbin\b\b\b\b";
char compare2[] = "not on bobbin";
char compare3[] = "have\b\b\b\b\b\b";
char compare4[] = "don't have";
char wrapedit[] = "wrapped";
char haveedit[] = "have";
char editing[9];
FILE *wrappedlist = fopen("../../wrapped", "r+");
FILE *havelist = fopen("../../havelist", "r+");
fseek(wrappedlist, i*14, SEEK_SET);
fseek(havelist, i*11, SEEK_SET);
printf("Edit? (y=yes, n=no)");
fgets(z, 2, stdin);
if ((*z=='y') && (strncmp(haveslist[i], compare4, (size_t)1) == 0)) {
printf("Switch \"don't have\" to \"have\"? (y=yes, n=no)");
fgets(y, 2, stdin);
if (*y=='y') {
fputs(compare3, havelist);
fclose(wrappedlist);
fclose(havelist);
return;
}
else if(*y=='n'){
fclose(wrappedlist);
fclose(havelist);
return;
}
printf("Invalid input.");
return;
}
else if ((*z=='y') && (strncmp(haveslist[i], compare3, (size_t)1) == 0)) {
fpurge(stdout);
printf("Edit \"wrapped\" or \"have\"?");
fpurge(stdin);
fgets(editing, 9, stdin);
len = strlen(editing);
editing[len-1]='\0';
if (strcmp(editing, wrapedit)==0) {
if (strncmp(wrapped[i], compare1, (size_t)1)==0) {
printf("Switch \"on bobbin\" to \"not on bobbin\"? (y=yes, n=no)");
fgets(y, 2, stdin);
if (*y=='y') {
fputs(compare2, wrappedlist);
fclose(wrappedlist);
fclose(havelist);
return;
}
else if(*y=='n'){
fclose(wrappedlist);
fclose(havelist);
return;
}
}
else if(strncmp(wrapped[i], compare2, (size_t)1)==0){
fpurge(stdout);
printf("Switch \"not on bobbin\" to \"on bobbin\"? (y=yes, n=no)");
fgets(y, 2, stdin);
if (*y=='y') {
fwrite(compare1, (size_t)strlen(compare1), 1, wrappedlist);
fclose(wrappedlist);
fclose(havelist);
return;
}
else if(*y=='n'){
fclose(wrappedlist);
fclose(havelist);
return;
}
fpurge(stdout);
printf("Invalid input.");
}
fpurge(stdout);
printf("You don't want to edit wrapped apparently.");
fclose(wrappedlist);
fclose(havelist);
return;
}
else if(strcmp(editing, haveedit)==0){
if (strncmp(haveslist[i], compare3, 1) == 0){
printf("Switch \"have\" to \"don't have\"? (y=yes, n=no)");
fgets(y, 2, stdin);
if (*y=='y') {
fputs(compare4, havelist);
fclose(wrappedlist);
fclose(havelist);
return;
}
else if(*y=='n'){
fclose(wrappedlist);
fclose(havelist);
return;
}
printf("Invalid input.");
}
else if(strncmp(haveslist[i], compare4, 1)==0){
printf("Switch \"don't have\" to \"have\"? (y=yes, n=no)");
fgets(y, 2, stdin);
if (*y=='y') {
fputs(compare3, havelist);
fclose(wrappedlist);
fclose(havelist);
return;
}
else if(*y=='n'){
fclose(wrappedlist);
fclose(havelist);
return;
}
}
printf("Invalid input.");
}
printf("Not editing.");
fclose(wrappedlist);
fclose(havelist);
return;
}
else if(*z=='n'){
fclose(wrappedlist);
fclose(havelist);
return;
}
printf("Invalid entry");
fclose(havelist);
fclose(wrappedlist);
return;
}
我可以在“编辑”后输入一个字符到fgets吗?提示,但后来我得到了exc_bad_access错误。请帮忙,谢谢.code:
答案 0 :(得分:7)
这是因为你的指针指向没有记忆。
char* z; // pointer points nowhere
/* snip */
fgets(z, 2, stdin); // writing to pointer that points nowhere: boom!
这会尝试将z
中的两个下一个字符放入stdin
。但是,z
指出无处可用。你需要它指向现有的内存:单独声明一个指针不足以获得它旁边的内存。
你可能想要一个2个字符的缓冲区:
char z[2];
使用此代码,z
将成为指向2个字符的足够内存的指针。 (在C中,可以在期望指针的任何地方传递数组。)
我的代码看起来并不太远,但y
会遇到同样的问题。
答案 1 :(得分:2)
您尚未为z
或y
分配任何内存,它们未被初始化,因此fgets
正在写入随机地址。