C语言- fscanf函数
C语言- fscanf函数
提问于 2017-11-09 21:28:29
为什么在下面的第一个fscanf函数中，%s的参数没有&但是%d的参数有&？
fscanf(fileR, "%s%d", meds[count].name, &meds[count].unitsInStock);
==> meds[count].name, &meds[count].unitsInStock
fscanf(fileR, "%i", &meds[count].sixmth[i]);
==> &meds[count].sixmth[i]回答 1
Stack Overflow用户
回答已采纳
发布于 2017-11-09 21:41:50
因为字符串已经是一个指针，我将在这个示例中解释它：
int * buff;
int n;
buff = &n;
//&n - address of n
//buff - address of n
//*buff - value of n
fscanf(stdin, "%d", &n); //it is only variable, it needs to pass address of it
fscanf(stdin, "%d", buff); //it is pointer to variable, it doesn't need to add &对于字符串(char*)，它是相同的：
char* str;
str = malloc(50);
fscanf(stdin, "%s", str); //it is poitner, there is already address页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/47211854
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