C 通过指针访问数组时出现分段错误
我将一个全局数组声明为C 通过指针访问数组时出现分段错误,c,arrays,pointers,operator-precedence,C,Arrays,Pointers,Operator Precedence,我将一个全局数组声明为 int Team1[12][8]; 当我调用函数时 int readLineupB(int teamNr){ int (*teamToRead)[12][8]; teamToRead = &Team1; for (int currentPlayersNumber = 1; currentPlayersNumber<11; currentPlayersNumber++) { for (int other
int Team1[12][8];
当我调用函数时
int readLineupB(int teamNr){
int (*teamToRead)[12][8];
teamToRead = &Team1;
for (int currentPlayersNumber = 1; currentPlayersNumber<11; currentPlayersNumber++) {
for (int otherNumber = 1; otherNumber<7; otherNumber++) {
*teamToRead[currentPlayersNumber][otherNumber] = 20;
}
}
return 1;
}
int readLineupB(int teamNr){
国际(*teamToRead)[12][8];
teamToRead=&Team1;
对于(int CurrentPlayerNumber=1;CurrentPlayerNumber请检查数据类型。
teamToRead
是指向int[12][8]
数组的指针
由于,下标运算符绑定的值高于取消引用的值
那么,万一
*teamToRead[currentPlayersNumber][otherNumber] = 20;
你想说的是
*(* ( teamToRead + currentPlayersNumber ) ) [otherNumber] = 20;
其中,指针算术变得非法,因为它们遵循指针类型,因此会越界
要解决这个问题,需要通过显式括号强制取消引用的优先级,如
(*teamToRead)[currentPlayersNumber][otherNumber] = 20;
另一种选择是放弃指向二维数组的指针,只使用指向(数组的第一个)一维数组的指针:
int Team1[12][8];
int readLineupB(int teamNr){
// declare teamToRead as a pointer to array of 8 ints
int (*teamToRead)[8];
// Team1 is of type int x[12][8] that readily decays
// to int (*x)[8]
teamToRead = Team1;
for (int currentPlayersNumber = 1; currentPlayersNumber<11; currentPlayersNumber++) {
for (int otherNumber = 1; otherNumber<7; otherNumber++) {
// no dereference here.
teamToRead[currentPlayersNumber][otherNumber] = 20;
}
}
return 0;
}
int Team1[12][8];
int readLineupB(int teamNr){
//将teamToRead声明为指向8整数数组的指针
int(*teamToRead)[8];
//Team1属于int x[12][8]类型,很容易衰变
//至整数(*x)[8]
teamToRead=Team1;
对于(int currentPlayersNumber=1;currentPlayersNumberhow about(*teamToRead)[currentPlayersNumber][otherNumber]=20;
?C数组从0
开始建立索引,然后上升到N-1
。@M.M已删除:D睡眠/咖啡太少。谢谢@SouravGhosh,括号解决了这个问题!关于数组从0开始的问题,我知道,我刚刚减小了“宽度”排除边缘错误的分配。我本不希望出现这种优先级,这当然可以解释错误。再次感谢。这种解决方案需要更少的书写,谢谢。但是,我更喜欢Sourav,因为它似乎更容易理解。人们会立即看到变量是指向二维数组的指针。