[英]C/Digital Logic - Why are my zero-initialized variables changing value?
They're actually initialized to the character '0' (ie 48).它们实际上被初始化为字符'0'(即48)。 I'm making a logic simulator and each bit is represented as a single char, either '0' or '1'.
我正在制作一个逻辑模拟器,每个位都表示为一个字符,“0”或“1”。 All of my other components have worked until I made an 8-bit selector.
在我制作了一个 8 位选择器之前,我所有的其他组件都可以正常工作。
This is the code for the 1-bit selector.这是 1 位选择器的代码。 It works fine.
它工作正常。
typedef char Bit;
...
typedef struct SelectorComp {
Bit store;
Bit data[2];
NandGate nand[3];
NotGate not;
Bit out;
} SelectorComp;
void initSelector(SelectorComp* selector) {
int i;
selector->store = '0';
for (i = 0; i < 2; i++) selector->data[i] = '0';
initNot(&selector->not);
for (i = 0; i < 3; i++) initNand(&selector->nand[i]);
Selector(selector);
}
void Selector(SelectorComp* selector) {
selector->nand[0].in[0] = selector->store;
selector->nand[0].in[1] = selector->data[0];
Nand(&selector->nand[0]);
selector->not.in = selector->store;
Not(&selector->not);
selector->nand[1].in[0] = selector->not.out;
selector->nand[1].in[1] = selector->data[1];
Nand(&selector->nand[1]);
selector->nand[2].in[0] = selector->nand[0].out;
selector->nand[2].in[1] = selector->nand[1].out;
Nand(&selector->nand[2]);
selector->out = selector->nand[2].out;
}
The 8-bit selector is basically 8 1-bit selectors tied together. 8 位选择器基本上是 8 个 1 位选择器捆绑在一起。
typedef struct Selector8Comp {
Bit store;
Bit data[2][8];
SelectorComp selector[8];
Bit out[8];
} Selector8Comp;
void initSelector8(Selector8Comp* selector8) {
int i, j;
selector8->store = '0';
for (i = 0; i < 2; i++) {
for (j = 0; j < 8; j++) {
selector8->data[i][j] = '0';
}
}
for (i = 0; i < 8; i++) initSelector(&selector8->selector[i]);
Selector8(selector8);
}
void Selector8(Selector8Comp* selector8) {
int i, j;
for (i = 0; i < 8; i++) {
selector8->selector[i].store = selector8->store;
for (j = 0; j < 2; j++) selector8->selector[i].data[j] = selector8->data[i][j];
Selector(&selector8->selector[i]);
selector8->out[i] = selector8->selector[i].out;
}
}
However, when I test it with the following function, I get unexpected output.但是,当我用下面的 function 测试它时,我得到了意想不到的 output。
void testSelector8() {
int i;
Selector8Comp* selector8 = (Selector8Comp*)malloc(sizeof(Selector8Comp));
initSelector8(selector8);
printf("Testing 8-bit selector\n");
printf("store: %c\n", selector8->store);
printf("byte 0: ");
for (i = 0; i < 8; i++) printf("%c", selector8->data[0][i]);
printf("\nbyte 1: ");
for (i = 0; i < 8; i++) printf("%c", selector8->data[1][i]);
printf("\nselectors:\n");
for (i = 0; i < 8; i++) {
printf("%d: store: %c in: %c%c out: %c\n", i, selector8->selector[i].store, selector8->selector[i].data[0], selector8->selector[i].data[1], selector8->selector[i].out);
}
printf("\noutput: ");
for (i = 0; i < 8; i++) printf("%c", selector8->out[i]);
printf("\n");
free(selector8);
}
Expected output:预期 output:
store: 0
byte 0: 00000000
byte 1: 00000000
selectors:
0: store: 0 in: 00 out: 0
1: store: 0 in: 00 out: 0
2: store: 0 in: 00 out: 0
3: store: 0 in: 00 out: 0
4: store: 0 in: 00 out: 0
5: store: 0 in: 00 out: 0
6: store: 0 in: 00 out: 0
7: store: 0 in: 00 out: 0
output: 0000000
Actual output:实际 output:
store: 0
byte 0: 00000000
byte 1: 00000000
selectors:
0: store: 0 in: 00 out: 0
1: store: 0 in: 00 out: 0
2: store: 0 in: 00 out: 0
3: store: 0 in: 11 out: 1
4: store: 0 in: 10 out: 0
5: store: 0 in: 10 out: 0
6: store: 0 in: 01 out: 1
7: store: 0 in: 01 out: 1
output: 00010011
Unless I'm reading my code wrong, all of those selectors should have their data initialized to "00".除非我读错了我的代码,否则所有这些选择器都应该将它们的数据初始化为“00”。 Why don't they?
他们为什么不呢?
EDIT: I'm using the latest gcc on ARM if that helps.编辑:如果有帮助,我在 ARM 上使用最新的 gcc。
I was reading my code wrong (imagine that).我读错了我的代码(想象一下)。 I switched up an i and aj (line 5 of the Selector8 function definition).
我切换了 i 和 aj(Selector8 function 定义的第 5 行)。 The j was supposed to come before the i in this case.
在这种情况下, j 应该在 i 之前。
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