struct instruction {
int value;
};
int n; // Number of instructions
struct instruction **instructions; //Required to use ** and dynamic array
假设我想存储n条指令并在每条指令中存储一个值。 我怎么用**说明呢? 所以我希望以后能够从特定指令中调用一个值。
非常感谢
到目前为止,我尝试了这些,一些scanfs和动态数组创建。 因此它需要多个计数器,然后获取线程数(pthreads),然后获取每个线程内的指令数。我试图找到一种方法来存储每个线程中的指令。 **给出了结构
int
main(void) {
scanf(" %d", &ncounters); //Ask for number of counters
int i;
if(counters = malloc(sizeof(struct counter)*ncounters)){
for( i=0; i < ncounters ;i++){
counters[i].counter = 0;
}
}
scanf(" %d", &nthreads); //Ask for number of threads
if(ninstructions = (int*) malloc(nthreads*sizeof(int))){
for( i=0; i < nthreads ;i++){
ninstructions[i] = 0;
}
}
for(i=0; i < nthreads ;i++){
scanf(" %d", &ninstructions[i]); //Ask for number of instructions within threads[i]
// Things got messy from here ...
instructions = malloc(sizeof(struct instruction*)*ninstructions[i]);
for(int j=0; j < ninstructions[j] ;j++){
instructions[j] = malloc(sizeof(struct instruction));
int x;
printf("Enter rep for thread %d inst %d.\n",i+1 ,j+1);
scanf(" %d", &x);
instructions[i][j].repetitions = x;
}
}
printf(" Instruction x: %d.\n", instructions[0][0].repetitions);
//=============================================================================
// Just testing with printf
printf("Number of counters: %d.\n", ncounters);
printf("Number of threads: %d.\n", nthreads);
for(i=0; i < nthreads; i++){
printf("Thread %d has %d instructions.\n", i+1, ninstructions[i]);
}
//=============================================================================
free(instructions);
free(ninstructions);
free(counters);
return 0;
}
我终于取得了一些进展,但在指令存储部分得到了分段错误。
struct counter *counters;
struct instruction{
struct counter *counter;
int repetitions;
void (*work_fn)(long long*);
};
for(i=0; i < nthreads ;i++){
scanf(" %d", &ninstructions[i]); //Ask for number of instructions within threads[i]
instructions = malloc(nthreads*sizeof(struct instruction *));
instructions[i] = malloc(ninstructions[i]*sizeof(struct instruction));
for(int j=0;j < ninstructions[i] ;j++){
int Srepetition;
char Sfunction;
int Scounter;
scanf(" %d %c %d", &Scounter, &Sfunction, &Srepetition);
//Problem seems to be here "Segmentation fault" ..............
instructions[i][j].repetitions = Srepetition;
instructions[i][j].counter = (counters+Scounter);
if(Sfunction == 'I'){
instructions[i][j].work_fn(&increment);
}else if(Sfunction == 'D'){
instructions[i][j].work_fn(&decrement);
}else if(Sfunction == '2'){
instructions[i][j].work_fn(&mult2);
}else{
printf("error\n");
}
printf("Thread: %d Instruction: %d.\n", i+1, j+1);
}
}
答案 0 :(得分:2)
只需一个*
足以创建一个动态数组,使用**
即可创建矩阵。
struct instruction *instructions = malloc(n * sizeof(struct instruction));
/* setting some random values */
for (int i=0;i<n;i++)
instructions[i]->value = i;
/* accessing values */
for (int i=0;i<n;i++)
printf("instruction %d value %d\n",i,instructions[i]->value);
/* don't forget to free */
free(instructions);
请查找有关dynamic arrays in C
的参考资料,以便进行更多调查。例如this one
修改强>
...如果你真的需要矩阵,这就是等效的代码:
int n,z; // for number cols and rows
struct instruction **instructions = malloc(n * sizeof(struct instruction *));
/* setting some random values */
for (int i=0;i<n;i++) {
instructions[i] = malloc(z * sizeof(struct instruction));
for (int j=0;j<z;j++)
instructions[i][j]->value = i * j;
}
/* accessing values */
for (int i=0;i<n;i++) {
for (int j=0;j<z;j++)
printf("instruction %d,%d value %d\n",i,j,instructions[i][j]->value);
}
/* don't forget to free */
for (int i=0;i<n;i++)
free(instructions[i]);
free(instructions);