c - 使用 scanf 修复一些警告

Question

int stringCheck(char*);

void compile(int *instructionSet2, int *accumulator2, int *instructionCounter2, int *instructionRegister2, int *operationCode2, int *operand2)
{

    char *readIn;
    readIn = (char*)malloc(sizeof(char)*11);

    while(fgets(readIn,11,stdin) != NULL)
    {
        *instructionCounter2 = scanf("%d",(int*)readIn);
        instructionSet2[*instructionCounter2] = (stringCheck((char*)scanf("%s",readIn)));
        instructionSet2[*instructionCounter2] += (scanf("%d",(int*)readIn));
    }
}

int stringCheck(char *stIn)
{
    if (strcmp(stIn,"READ") == 0)
    {
        return 10 * 100;
    } /* Snipped to just give an idea of the numeric value assignment for words read. */
}

好吧，我在这里想做的是弄清楚如何正确阅读如下一组指令:

00 READ 9
01 READ 10
02 LOAD 9
03 SUB 10
04 BRNG 7
05 WRIT 9
06 HALT 99
07 WRIT 10
08 HALT 99
09 SET 0
10 SET 0

虽然我在尝试阅读文字时遇到了一些困难。基本上，我只想扫描一个 int、一个字符串，然后扫描另一个 int，并在未到达文件末尾时继续执行此操作。

事实证明，scanf 返回成功读取的字符数而不是指针，因此 while 循环内的第二行在编译时警告我这一点。

我正在尝试特别修复此警告:

virtualcomp.c:66:56: warning: cast to pointer from integer of different size [-Wint-to-pointer-cast]

专门指向这一行:

instructionSet2[*instructionCounter2] = (stringCheck((char*)scanf("%s",readIn)));

我该如何解决这个问题？

注意:我试图避免输入更多变量，仅使用现有的变量。

Answer 1

您可以通过一次 scanf 调用来解析所有三个变量。注意 scanf 返回分配的输入变量的数量。我认为你需要这样的东西:

int n1, n2;
char buf[12];
...
...
sscanf(readIn, "%d %s %d",&n1, buff, &n2);

*instructionCounter2 = n1;
instructionSet2[*instructionCounter2] = stringCheck(buff);
instructionSet2[*instructionCounter2] += n2;