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在下面的示例中，我想了解为什么将 4195520 的值添加到 chars[2] （计算空格）。

char input;
int chars[4];  // [0] = newlines; [1] = tabs; [2] = spaces; [3] = all_else 
int i, j;

printf("--\n");
printf("please enter input\n");
printf("~[enter] to end\n\n");
while ((input = getchar()) != EOF) {
    if (input == '\n') {++chars[0];}
    else if (input == '\t') {++chars[1];}
    else if (input == ' ') {++chars[2];}  // 4195520 is added (plus the amount of actual spaces), but only if I add to an array member
                                          // if I increment an integer variable, it counts correctly
    else if (input == '~') {printf("\n"); break;}
    else {++chars[3];}
}

printf("--\n");
printf("the frequency of different characters\n");
printf("each dot represents one occurance\n");
for (i=0; i<=3; ++i) {
    if (i == 0) {printf("newlines:\t");}
    else if (i == 1) {printf("tabs:\t\t");}
    else if (i == 2) {printf("spaces:\t\t");}
    else if (i == 3) {printf("all else:\t");}
    for (j=0; j<chars[i]; ++j) {printf(".");}  // ~4.2 million dots are printed for spaces
    printf("\n");
}

正确计数的解决方案是将数组成员初始化为 0，但我想了解为什么只有空格以及为什么只使用数组成员作为计数器，每次都会递增到数字 4195520。

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1 回答 1

3

It's garbage value: some value that just happens to be at the particular memory location because of a previous operation, and has not been cleared. You might be getting 4195520 this time around, but if you run some other program and then re-run your code you'll probably see something different. You'll almost certainly see something different on a different machine.

As you note, the correct way to do the counting is this:

int chars[4] = {0};
于 2013-08-26T17:48:24.553 回答