C语言 错误：赋值使指针从整数而不进行强制转换
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Error:Assignment makes pointer from integer without a cast
提问by Elia
int findNumber(char *exp,int i,int *num)
{
int k=i;
char *p;
p=exp[i]; //<-- here
while(*p>='0'&&*p<='9')
{
(*num)=(*num)*10+(*p);
k++;
p++;
}
return k;
}
i keep getting that error in line: (p=exp[i];) Im trying to send a char array, and (i,num) integers, the 'i' im just putting it to be 0 for now, until the code works so dont give attention to it. but the function should return the place of the first character in "exp" that is not a number, with being sure that all the ones before are numbers.
我不断收到该错误：(p=exp[i];) 我正在尝试发送一个字符数组和 (i,num) 整数，'i' 我只是暂时将它设置为 0，直到代码作品所以不要关注它。但是该函数应该返回“exp”中第一个不是数字的字符的位置，并确保之前的所有字符都是数字。
回答by simonc
pis a char*so you need to assign a pointer to it but exp[i]returns a single charelement from an array. Try
p是 achar*所以你需要为它分配一个指针，但从数组中exp[i]返回一个char元素。尝试
p = &exp[i];
or
或者
p = (exp + i);
instead.
反而。
回答by hmatar
do correct the line to correspond to the following p=&(exp[i]);this means that you are assigning the pointer of exp[i] to p.
更正该行以对应于以下内容，p=&(exp[i]);这意味着您将 exp[i] 的指针分配给 p。
回答by Hyman
You need take the address-of instead of and then pass to p, that's a pointer. In other words,you are assign charto a char* returned from value atiindex inexp` array.
您需要使用 address-of 而不是，然后传递给p，这是一个指针。换句话说，你被分配 char给一个char* returned from value ati index inexp` 数组。
Try this: p = &exp[i];
尝试这个： p = &exp[i];
