在C中将值分配给单独的数组后，要更改数组的第0个值

Question

我正在读取文件

（（标记为in1的文件内容：]

8
0 2 1 4 2 2 3 1
3 2 5 4 6 7 9 2 8  

通过命令行传递到我的程序。

（（这里是最低工作版本）：

/*
 *Run using make with accompanying Makefile using file in1 as input on command line
 *
 * */
#include<stdio.h>
#include<stdlib.h>

int main(int argc, char* argv[]){

if(argc != 2){

   fprintf(stderr,
      "Error: Improper amount of command line arguments. Need one input file.\n");
   exit(EXIT_FAILURE);

}

FILE* fp = fopen(argv[1], "r");

int n;
int* A = calloc(n+1, sizeof(int));
int* idx = calloc(n, sizeof(int));

fscanf(fp, "%d", &n); //assigns n from file

printf("N is equal to : %d\n", n);

for(int i = 0; i < n; i++){ //assigns operations from file

fscanf(fp, "%d", &idx[i]);

}

for(int i = 0; i < n; i++){

   printf("idx[%d] = %d\n", i, idx[i]);

}

for(int i = 0; i <= n; i++){//set operands

   fscanf(fp, "%d", &A[i]);

}
for(int i = 0; i <= n; i++){

   printf("A[%d] = %d\n", i, A[i]);

}

for(int i = 0; i < n; i++){

   printf("idx[%d] = %d\n", i, idx[i]);

}
}

此最小版本仍会产生相同的结果。这是输出：

N is equal to : 8
idx[0] = 0
idx[1] = 2
idx[2] = 1
idx[3] = 4
idx[4] = 2
idx[5] = 2
idx[6] = 3
idx[7] = 1
A[0] = 3
A[1] = 2
A[2] = 5
A[3] = 4
A[4] = 6
A[5] = 7
A[6] = 9
A[7] = 2
A[8] = 8
idx[0] = 8
idx[1] = 2
idx[2] = 1
idx[3] = 4
idx[4] = 2
idx[5] = 2
idx[6] = 3
idx[7] = 1

请参阅idx [0]如何首先等于0，然后在分配A之后，其值为8？

这是为什么？

（使文件运行程序）：

PROGRAM_NAME = minRecreate
FLAGS        = -std=c99 -Wall
SOURCE       = $(PROGRAM_NAME).c
OBJECT       = $(PROGRAM_NAME).o
EXEBIN       = $(PROGRAM_NAME)

$(EXEBIN) : $(OBJECT)
        gcc -o $(EXEBIN) $(OBJECT)

$(OBJECT) : $(SOURCE)
        gcc -c $(FLAGS) $(SOURCE)

clean :
        rm $(EXEBIN) $(OBJECT)

memcheck : $(EXEBIN)
        valgrind --leak-check=full $(EXEBIN) in1

Answer 1

第一次使用n时，未初始化。程序的其余部分导致未定义的行为，但是最有可能发生的是A不够大，因此最终导致了idx崩溃。