不使用 &scanf() 中的变量是语法错误还是运行时错误? [英] Is not using & with a variable in scanf() a syntax error or a runtime error?

问题描述

有人告诉我语法错误来自一个错误并且不执行程序.但是，运行时错误仍会启动程序，但会在中间崩溃.如果我在调用 scanf() 时不小心没有将 & 放在变量前面，那会被认为是语法错误还是运行时错误?

I was taught that a syntax error comes from a single mistake and doesn't execute the program. However, a runtime error still starts the program but crashes in the middle. If I accidentally didn't put & in front of a variable in a call to scanf(), would that considered to be a syntax error or a runtime error?

例如

int main(void)
{
    int a;
    printf("input a integer number >>");
    scanf("%d\n", a);
    printf("the input number is %d\n", a);
    return 0;
}

在 scanf 语句中，我在 a 前面没有 & 这样会崩溃，但它仍然执行第一个 printf 语句，所以它会显示一些打印的文本.

In the scanf statement, I don't have & in front of a so that would crash, but it still executes the first printf statement, so it would show some printed text.

它会同时是运行时错误和语法错误吗?

Is it going to be both a runtime error and a syntax error?

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