错误：不兼容的类型分配给键入时'的char [25]'从类型'字符*' [英] Error: incompatible types when assigning to type ‘char[25]’ from type ‘char *’

问题描述

我试图分配结构到地图的价值，但在编译后会出现以下错误信息：

 错误：不兼容的类型分配给键入时'的char [25]'从类型'字符*'
 

在

 地图[I] .N = M.N
 

我的结构的定义是这样的：

 结构M1 {
INT℃;
CHAR N [25];
INT q_m;
INT q;};
 

我的code的部分：结果

 结构M1米;
    结构M1 *地图= 0;
    scanf函数（％d个，＆安培; M.C）;
    scanf函数（％S，＆安培; M.N）;
    scanf函数（％d个，＆安培; m.q_m）;
    scanf函数（％d个，＆安培; m.q）;    地图[I] .C = M.C;
    地图[I] .N = M.N;
    地图[I] .q_m = m.q_m;
    地图[I] .Q = m.q;
 

解决方案

阵列前pressions可能不是一个赋值的目标;在 = 运营商没有定义一个数组的内容的复制到其他。

如果 N 意味着持有0结尾的字符串，用的strcpy ：

 的strcpy（图[I] .N，M.N）;
 

如果 N 是为了举行一个非0结尾的字符串（或字符嵌入0值的序列），可以使用的memcpy ：

 的memcpy（图[I] .N，M.N，sizeof的映射[I] .N）;
 

除非它是的sizeof 或一元＆放大器的操作; 运营商，或者是一个字符串使用初始化在声明中另一个数组，键入 T N个元素的数组的前pression将被转换（衰变），以一个前$ p类型$ pssion指针 T ，和前pression的值将是第一个元素的地址。

这就是为什么你得到了你做了错误信息;恩pression M.N 有型字符 25个元素的数组因为它不是的操作数的的sizeof 或一元＆安培; 运营商，它被转换为类型的char * 。 地图[I] .N 未转换（它停留键入的char [25] ），但正如我所说此前，前阵pressions可能不是赋值运算符的目标。

I'm trying to assign the values of a struct to a map but the following error message appears after compiling:

error: incompatible types when assigning to type ‘char[25]’ from type ‘char *’ 

in

map[i].n=m.n

My struct is defined this way:

struct m1{
int c;
char n[25];
int q_m;
int q;};

Part of my code:

    struct m1 m;
    struct m1 *map = 0;
    scanf("%d",&m.c);
    scanf("%s",&m.n);
    scanf("%d",&m.q_m);
    scanf("%d",&m.q);

    map[i].c=m.c;
    map[i].n=m.n;
    map[i].q_m=m.q_m;
    map[i].q=m.q;

解决方案

Array expressions may not be the target of an assignment; the = operator isn't defined to copy the contents of one array to the other.

If n is meant to hold a 0-terminated string, use strcpy:

strcpy( map[i].n, m.n );

If n is meant to hold a non-0-terminated string (or a sequence of characters with embedded 0 values), use memcpy:

memcpy( map[i].n, m.n, sizeof map[i].n );

Unless it is the operand of the sizeof or unary & operators, or is a string literal being used to initialize another array in a declaration, an expression of type "N-element array of T" will be converted ("decay") to an expression of type "pointer to T", and the value of the expression will be the address of the first element.

That's why you got the error message you did; the expression m.n has type "25-element array of char"; since it wasn't the operand of the sizeof or unary & operators, it was converted to type char *. map[i].n wasn't converted (it stayed type char [25]), but as I said earlier, array expressions may not be the target of the assignment operator.

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