C语言 赋值从指针生成整数而无需强制转换 [-Wint-conversion
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Assignment makes integer from pointer without a cast [-Wint-conversion
提问by Addon
I really don't understand why I have such error knowing that tmpand keyare the same type and size.
我真的不明白为什么我知道会出现这样的错误tmp并且key是相同的类型和大小。
uint8_t key[8] = {0x00,0x01,0x02,0x03,0x04,0x05,0x06,0x07};
void change() {
int i;
uint8_t *tmp[8];
for(i=0; i<8; i++){
tmp[i] = key[(i+3)%8];
}
}
This produces:
这产生：
warning: assignment makes integer from pointer without a cast [-Wint-conversion
警告：赋值从指针生成整数而无需强制转换 [-Wint-conversion
回答by Sourav Ghosh
tmpandkeyare the same type
tmp并且key是相同的类型
NO. They are not. They both are arrays, but the datatype is different. One is a uint8_t *array, another is a uint8_tarray.
否。他们不是。它们都是数组，但数据类型不同。一个是uint8_t *数组，另一个是uint8_t数组。
Change
改变
uint8_t *tmp[8];
to
到
uint8_t tmp[8] = {0};
回答by mksteve
Not clear what you want here but if you want tmp[x]to reflect the value in key[y]then
尚不清楚你想要的这里，但如果你想tmp[x]以反映价值key[y]，然后
tmp[i] = &key[(i+3)%8]; /* tmp[i] now points at key[ (i+3)%8];
// key[3] = 5; /* These two lines modify the same memory */
// (*tmp[0]) = 5; /* */
Otherwise if you want tmp to be separate, then ...
否则，如果您希望 tmp 分开，那么...
uint8_t tmp[8]; /* change type to be non-pointer. */
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